## Trig. Equations Examples using CAST Diagrams

These lessons, with videos, examples and step-by-step solutions help A Level Maths students learn to solve trigonometric problems.

Related Pages Trigonometric Functions Trigonometric Graphs Trigonometric Identities Lessons On Trigonometry More Lessons for A Level Maths

What is the CAST diagram? The Cast diagram helps us to remember the signs of the trigonometric functions in each of the quadrants. The CAST diagram is also called the Quadrant Rule or the ASTC diagram.

In the first quadrant, the values are all positive. In the second quadrant, only the values for sin are positive. In the third quadrant, only the values for tan are positive. In the fourth quadrant, only the values for cos are positive.

Note that the mnemonic CAST goes anticlockwise starting from the 4th quadrant.

The following diagram shows how the CAST diagram or Quadrant rule can be used. Scroll down the page for more examples and solutions.

Using the Quadrant Rule to solve trig. equations

Quadrant Rule for solving trig equations with different ranges

Core (2) Graphs of Trigonometric Functions (4) - CAST Diagram or Unit Circle

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## 6.4 Trigonometric equations

6.4 trigonometric equations (embhm).

Solving trigonometric equations requires that we find the value of the angles that satisfy the equation. If a specific interval for the solution is given, then we need only find the value of the angles within the given interval that satisfy the equation. If no interval is given, then we need to find the general solution. The periodic nature of trigonometric functions means that there are many values that satisfy a given equation, as shown in the diagram below.

## Worked example 12: Solving trigonometric equations

Solve for $$\theta$$ (correct to one decimal place), given $$\tan \theta = 5$$ and $$\theta \in [\text{0}\text{°};\text{360}\text{°}]$$.

## Use a calculator to solve for $$\theta$$

This value of $$\theta$$ is an acute angle which lies in the first quadrant and is called the reference angle.

## Use the CAST diagram to determine in which quadrants $$\tan \theta$$ is positive

The CAST diagram indicates that $$\tan \theta$$ is positive in the first and third quadrants, therefore we must determine the value of $$\theta$$ such that $$\text{180}\text{°} < \theta < \text{270}\text{°}$$.

Using reduction formulae, we know that $$\tan (\text{180}\text{°} + \theta) = \tan \theta$$

## Use a calculator to check that the solution satisfies the original equation

$$\theta = \text{78,7}\text{°}$$ or $$\theta = \text{258,7}\text{°}$$.

## Worked example 13: Solving trigonometric equations

Solve for $$\alpha$$ (correct to one decimal place), given $$\cos \alpha = -\text{0,7}$$ and $$\theta \in [\text{0}\text{°};\text{360}\text{°}]$$.

## Use a calculator to find the reference angle

To determine the reference angle, we do not include the negative sign. The reference angle must be an acute angle in the first quadrant, where all the trigonometric functions are positive.

## Use the CAST diagram to determine in which quadrants $$\cos \alpha$$ is negative

The CAST diagram indicates that $$\cos \alpha$$ is negative in the second and third quadrants, therefore we must determine the value of $$\alpha$$ such that $$\text{90}\text{°} < \alpha < \text{270}\text{°}$$.

Using reduction formulae, we know that $$\cos (\text{180}\text{°} - \alpha) = -\cos \alpha$$ and $$\cos (\text{180}\text{°} + \alpha) = -\cos \alpha$$

Note: the reference angle $$(\text{45,6}\text{°})$$ does not form part of the solution.

$$\alpha = \text{134,4}\text{°}$$ or $$\alpha = \text{225,6}\text{°}$$.

## Worked example 14: Solving trigonometric equations

Solve for $$\beta$$ (correct to one decimal place), given $$\sin \beta = -\text{0,5}$$ and $$\beta \in [-\text{360}\text{°};\text{360}\text{°}]$$.

To determine the reference angle, we use a positive value.

## Use the CAST diagram to determine in which quadrants $$\sin \beta$$ is negative

The CAST diagram indicates that $$\sin \beta$$ is negative in the third and fourth quadrants. We also need to find the values of $$\beta$$ such that $$-\text{360}\text{°} \leq \beta \leq \text{360}\text{°}$$.

Using reduction formulae, we know that $$\sin (\text{180}\text{°} + \beta) = - \sin \beta$$ and $$\sin (\text{360}\text{°} - \beta) = - \sin \beta$$

Notice: the reference angle $$(\text{30}\text{°})$$ does not form part of the solution.

$$\beta = -\text{150}\text{°}$$, $$-\text{30}$$$$\text{°}$$, $$\text{210}$$$$\text{°}$$ or $$\text{330}$$$$\text{°}$$.

## Solving trigonometric equations

Determine the values of $$\alpha$$ for $$\alpha \in [\text{0}\text{°};\text{360}\text{°}]$$ if:

$$4 \cos \alpha = 2$$

$$\alpha = \text{60}\text{°}; \text{300}\text{°}$$

$$\sin \alpha + \text{3,65} = 3$$

$$\alpha = \text{220,5}\text{°}; \text{319,5}\text{°}$$

$$\tan \alpha = 5\frac{1}{4}$$

$$\alpha = \text{79,2}\text{°}; \text{259,2}\text{°}$$

$$\cos \alpha + \text{0,939} = 0$$

$$\alpha = \text{200,1}\text{°}; \text{339,9}\text{°}$$

$$5 \sin \alpha = 3$$

$$\alpha = \text{36,9}\text{°}; \text{143,1}\text{°}$$

$$\frac{1}{2} \tan \alpha = -\text{1,4}$$

$$\alpha = \text{109,7}\text{°}; \text{289,7}\text{°}$$

Determine the values of $$\theta$$ for $$\theta \in [-\text{360}\text{°};\text{360}\text{°}]$$ if:

$$\sin \theta = \text{0,6}$$

Negative angles:

$$\theta = -\text{323,1}\text{°}; -\text{216,9}\text{°}; \text{36,9}\text{°}; \text{143,1}\text{°}$$

$$\cos \theta + \frac{3}{4} = 0$$

$$\theta = -\text{221,4}\text{°}; -\text{138,6}\text{°}; \text{138,6}\text{°}; \text{221,4}\text{°}$$

$$3 \tan \theta = 20$$

$$\theta = -\text{278,5}\text{°}; -\text{98,5}\text{°}; \text{81,5}\text{°}; \text{261,5}\text{°}$$

$$\sin \theta = \cos \text{180}\text{°}$$

$$\theta = -\text{90}\text{°}; \text{270}\text{°}$$

$$2 \cos \theta = \frac{4}{5}$$

$$\theta = -\text{293,6}\text{°}; -\text{66,4}\text{°}; \text{66,4}\text{°}; \text{293,6}\text{°}$$

## The general solution (EMBHN)

In the previous worked example, the solution was restricted to a certain interval. However, the periodicity of the trigonometric functions means that there are an infinite number of positive and negative angles that satisfy an equation. If we do not restrict the solution, then we need to determine the general solution to the equation. We know that the sine and cosine functions have a period of $$\text{360}$$$$\text{°}$$ and the tangent function has a period of $$\text{180}$$$$\text{°}$$.

Method for finding the general solution:

• Determine the reference angle (use a positive value).
• Use the CAST diagram to determine where the function is positive or negative (depending on the given equation).
• Find the angles in the interval $$[\text{0}\text{°}; \text{360}\text{°}]$$ that satisfy the equation and add multiples of the period to each answer.
• Check answers using a calculator.

## Worked example 15: Finding the general solution

Determine the general solution for $$\sin \theta = \text{0,3}$$ (give answers correct to one decimal place).

## Use CAST diagram to determine in which quadrants $$\sin \theta$$ is positive

The CAST diagram indicates that $$\sin \theta$$ is positive in the first and second quadrants.

Using reduction formulae, we know that $$\sin (\text{180}\text{°} - \theta) = \sin \theta$$.

where $$k \in \mathbb{Z}$$.

## Check that the solution satisfies the original equation

We can select random values of $$k$$ to check that the answers satisfy the original equation.

Let $$k = 4$$:

This solution is correct.

Similarly, if we let $$k = -2$$:

This solution is also correct.

$$\theta = \text{17,5}\text{°} + k \cdot \text{360}\text{°}$$ or $$\theta = \text{162,5}\text{°} + k \cdot \text{360}\text{°}$$.

## Worked example 16: Finding the general solution

Determine the general solution for $$\cos 2\theta = - \text{0,6427}$$ (give answers correct to one decimal place).

## Use CAST diagram to determine in which quadrants $$\cos \theta$$ is negative

The CAST diagram shows that $$\cos \theta$$ is negative in the second and third quadrants.

Therefore we use the reduction formulae $$\cos (\text{180}\text{°} - \theta) = - \cos \theta$$ and $$\cos (\text{180}\text{°} + \theta) = - \cos \theta$$.

Remember: also divide the period $$(\text{360}\text{°})$$ by the coefficient of $$\theta$$.

Let $$k = 2$$:

Similarly, if we let $$k = -5$$:

$$\theta = \text{65}\text{°} + k \cdot \text{180}\text{°}$$ or $$\theta = \text{115}\text{°} + k \cdot \text{180}\text{°}$$.

## Worked example 17: Finding the general solution

Determine the general solution for $$\tan (2\alpha - \text{10}\text{°}) = \text{2,5}$$ such that $$-\text{180}\text{°} \leq \alpha \leq \text{180}\text{°}$$ (give answers correct to one decimal place).

## Make a substitution

To solve this equation, it can be useful to make a substitution: let $$x = 2\alpha - \text{10}\text{°}$$.

## Use CAST diagram to determine in which quadrants the tangent function is positive

We see that $$\tan x$$ is positive in the first and third quadrants, so we use the reduction formula $$\tan (\text{180}\text{°} + x) = \tan x$$. It is also important to remember that the period of the tangent function is $$\text{180}$$$$\text{°}$$.

Remember: to divide the period $$(\text{180}\text{°})$$ by the coefficient of $$\alpha$$.

## Find the answers within the given interval

Substitute suitable values of $$k$$ to determine the values of $$\alpha$$ that lie within the interval ($$-\text{180}\text{°} \leq \alpha \leq \text{180}\text{°}$$).

Notice how some of the values repeat. This is because of the periodic nature of the tangent function. Therefore we need only determine the solution:

$\alpha = \text{39,1}\text{°} + k \cdot \text{90}\text{°}$ for $$k \in \mathbb{Z}$$.

$$\alpha = -\text{140,9}\text{°}$$; $$-\text{50,9}$$$$\text{°}$$; $$\text{39,1}$$$$\text{°}$$ or $$\text{129,1}$$$$\text{°}$$.

## Worked example 18: Finding the general solution using co-functions

Determine the general solution for $$\sin (\theta - \text{20}\text{°}) = \cos 2\theta$$.

## Use the CAST diagram to decide on the quadrants

The right hand side of the equation is positive, and the sine function is positive in the first and second quadrants. Therefore, we will reduce the equation as if the solution were in the first and second quadrants (using the angle and $$\text{180}\text{°}$$ minus the angle).

## Solve the equation

Solution 1:

Solution 2:

For the first solution:

For the second solution:

$$\theta = \text{36,67}\text{°} + k \cdot \text{120}\text{°}$$ or $$\theta = \text{250}\text{°} + k \cdot \text{360}\text{°}, \quad k \in \mathbb{Z}$$

## General solution

Find the general solution for each equation.

• Hence, find all the solutions in the interval $$[-\text{180}\text{°};\text{180}\text{°}]$$.

$$\cos (\theta + \text{25}\text{°}) = \text{0,231}$$

$$\theta = -\text{128,36}\text{°}; -\text{101,64}\text{°}; \text{51,64}\text{°}$$

$$\sin 2\alpha = -\text{0,327}$$

$$\theta = -\text{80,45}\text{°}; -\text{9,54}\text{°}; \text{99,55}\text{°}; \text{170,46}\text{°}$$

$$2 \tan \beta = -\text{2,68}$$

$$\theta = -\text{53,27}\text{°}; \text{126,73}\text{°}$$

$$\cos \alpha = 1$$

$$\alpha = \text{0}\text{°}$$

$$4 \sin \theta = 0$$

$$\theta = -\text{180}\text{°}; \text{0}\text{°}; \text{180}\text{°}$$

$$\cos \theta = -1$$

$$\theta = -\text{180}\text{°}; \text{180}\text{°}$$

$$\tan \frac{\theta}{2} = \text{0,9}$$

$$4 \cos \theta +3 = 1$$

$$\theta = -\text{120}\text{°}; \text{120}\text{°}$$

$$\sin 2\theta = -\frac{\sqrt{3}}{2}$$

$$\theta = -\text{60}\text{°}; -\text{30}\text{°}; \text{120}\text{°}; \text{150}\text{°}$$

$$\cos (\theta + \text{20}\text{°}) = 0$$

$$\theta = - \text{20}\text{°} + n \cdot \text{360}\text{°}$$

$$\sin 3\alpha = -1$$

$$\alpha = \text{30}\text{°} + n \cdot \text{120}\text{°}$$

$$\tan 4\beta = \text{0,866}$$

$$\beta = \text{10,25}\text{°} + n \cdot \text{45}\text{°} \text{ or }\beta = \text{55,25}\text{°} + n \cdot \text{45}\text{°}$$

$$\cos (\alpha - \text{25}\text{°}) = \text{0,707}$$

$$\alpha = \text{70}\text{°} + n \cdot \text{360}\text{°} \text{ or }\alpha = \text{340}\text{°} + n \cdot \text{360}\text{°}$$

$$2 \sin \frac{3\theta}{2} = -1$$

$$\theta = \text{140}\text{°} + n \cdot \text{240}\text{°} \text{ or }\theta = \text{220}\text{°} + n \cdot \text{240}\text{°}$$

$$5 \tan (\beta + \text{15}\text{°}) = \frac{5}{\sqrt{3}}$$

We can use our knowledge of algebraic equations to solve quadratic trigonometric equations.

## Worked example 19: Quadratic trigonometric equations

Find the general solution of $$4 \sin^2 \theta = 3$$.

## Simplify the equation and determine the reference angle

Determine in which quadrants the sine function is positive and negative.

The CAST diagram shows that $$\sin \theta$$ is positive in the first and second quadrants and negative in the third and fourth quadrants.

Positive in the first and second quadrants:

Negative in the third and fourth quadrants:

$$\theta = \text{60}\text{°} + k \cdot \text{360}\text{°}$$ or $$\text{120}\text{°} + k \cdot \text{360}\text{°}$$ or $$\text{240}\text{°} + k \cdot \text{360}\text{°}$$ or $$\text{300}\text{°} + k \cdot \text{360}\text{°}$$

## Worked example 20: Quadratic trigonometric equations

Find $$\theta$$ if $$2 \cos^2 \theta - \cos \theta - 1 = 0$$ for $$\theta \in [-\text{180}\text{°};\text{180}\text{°}]$$.

## Factorise the equation

Simplify the equations and solve for $$\theta$$, substitute suitable values of $$k$$.

Determine the values of $$\theta$$ that lie within the the given interval $$\theta \in [-\text{180}\text{°};\text{180}\text{°}]$$ by substituting suitable values of $$k$$.

If $$k = -1$$,

If $$k = 0$$,

If $$k = 1$$,

## Alternative method: substitution

We can simplify the given equation by letting $$y = \cos \theta$$ and then factorising as:

We substitute $$y = \cos \theta$$ back into these two equations and solve for $$\theta$$.

$$\theta = -\text{120}\text{°}$$; $$\text{0}$$$$\text{°}$$; $$\text{120}$$$$\text{°}$$

## Worked example 21: Quadratic trigonometric equations

Find $$\alpha$$ if $$2 \sin^2 \alpha - \sin \alpha \cos \alpha = 0$$ for $$\alpha \in [\text{0}\text{°};\text{360}\text{°}]$$.

## Factorise the equation by taking out a common factor

Simplify the equations and solve for $$\alpha$$.

To simplify further, we divide both sides of the equation by $$\cos \alpha$$.

Determine the values of $$\alpha$$ that lie within the the given interval $$\alpha \in [\text{0}\text{°}; \text{360}\text{°}]$$ by substituting suitable values of $$k$$.

If $$k = 0$$:

If $$k = 1$$:

If $$k = 2$$:

$$\alpha = \text{0}\text{°}$$; $$\text{26,6}$$$$\text{°}$$; $$\text{180}$$$$\text{°}$$; $$\text{206,6}$$$$\text{°}$$; $$\text{360}$$$$\text{°}$$

Find the general solution for each of the following equations:

$$\cos 2\theta = 0$$

$$\theta = \text{45}\text{°} + k \cdot \text{180}\text{°}$$ or $$\theta = \text{135}\text{°} + k \cdot \text{180}\text{°}$$

$$\sin (\alpha + \text{10}\text{°}) = \frac{\sqrt{3}}{2}$$

$$\alpha = \text{50}\text{°} + k \cdot \text{360}\text{°}$$ or $$\alpha = \text{110}\text{°} + k \cdot \text{360}\text{°}$$

$$2 \cos \frac{\theta}{2} - \sqrt{3} = 0$$

$$\theta = \text{60}\text{°} + k \cdot \text{720}\text{°}$$ or $$\theta = \text{660}\text{°} + k \cdot \text{720}\text{°}$$

$$\frac{1}{2} \tan (\beta - \text{30}\text{°}) = -1$$

$$\beta = \text{146,6}\text{°} + k \cdot \text{180}\text{°}$$

$$5 \cos \theta = \tan \text{300}\text{°}$$

$$\theta = \text{110,27}\text{°} + k \cdot \text{360}\text{°}$$ or $$\theta = \text{249,73}\text{°} + k \cdot \text{360}\text{°}$$

$$3 \sin \alpha = -\text{1,5}$$

$$\alpha = \text{210}\text{°} + k \cdot \text{360}\text{°}$$ or $$\alpha = \text{330}\text{°} + k \cdot \text{360}\text{°}$$

$$\sin 2 \beta = \cos (\beta + \text{20}\text{°})$$

$$\beta = \text{23,3}\text{°} + k \cdot \text{120}\text{°}$$

$$\text{0,5} \tan \theta + \text{2,5}= \text{1,7}$$

$$\theta = \text{122}\text{°} + k \cdot \text{180}\text{°}$$

$$\sin (3 \alpha - \text{10}\text{°}) = \sin (\alpha + \text{32}\text{°})$$

$$\alpha = \text{21}\text{°} + k \cdot \text{180}\text{°}$$ or $$\alpha = \text{39,5}\text{°} + k \cdot \text{90}\text{°}$$

$$\sin 2 \beta = \cos 2 \beta$$

$$\beta = \text{22,5}\text{°} + k \cdot \text{90}\text{°}$$

Find $$\theta$$ if $$\sin^2 \theta + \frac{1}{2} \sin \theta = 0$$ for $$\theta \in [\text{0}\text{°};\text{360}\text{°}]$$.

$$\theta = \text{0}\text{°}, \text{180}\text{°}, \text{210}\text{°}, \text{330}\text{°} \text{ or } \text{360}\text{°}$$

Determine the general solution for each of the following:

$$2 \cos^2 \theta - 3 \cos \theta = 2$$

$$\theta = \text{120}\text{°} + k \cdot \text{360}\text{°}$$ or $$\theta = \text{240}\text{°} + k \cdot \text{360}\text{°}$$

$$3 \tan^2 \theta + 2 \tan \theta = 0$$

$$\theta = \text{0}\text{°} + k \cdot \text{180}\text{°}$$ or $$\theta = \text{146,3}\text{°} + k \cdot \text{180}\text{°}$$

$$\cos^2 \alpha = \text{0,64}$$

$$\alpha = \text{36,9}\text{°} + k \cdot \text{360}\text{°}$$ or $$\alpha = \text{143,1}\text{°} + k \cdot \text{360}\text{°}$$ or $$\alpha = \text{216,9}\text{°} + k \cdot \text{360}\text{°}$$ or $$\alpha = \text{323,1}\text{°} + k \cdot \text{360}\text{°}$$

$$\sin (4\beta + \text{35}\text{°}) = \cos (\text{10}\text{°} - \beta)$$

$$\beta = \text{15}\text{°} + k \cdot \text{120}\text{°}$$ or $$\beta = \text{75}\text{°} + k \cdot \text{120}\text{°}$$

$$\sin (\alpha + \text{15}\text{°}) = 2 \cos (\alpha + \text{15}\text{°})$$

$$\alpha = \text{48,4}\text{°} + k \cdot \text{180}\text{°}$$

$$\sin^2 \theta - 4\cos^2 \theta = 0$$

$$\theta = \text{63,4}\text{°} + k \cdot \text{180}\text{°}$$ or $$\theta = \text{116,6}\text{°} + k \cdot \text{180}\text{°}$$

$$\dfrac{\cos (2 \theta + \text{30}\text{°})}{2} + \text{0,38} = 0$$

$$\theta = \text{54,8}\text{°} + k \cdot \text{180}\text{°}$$ or $$\theta = \text{95,25}\text{°} + k \cdot \text{180}\text{°}$$

Find $$\beta$$ if $$\frac{1}{3} \tan \beta = \cos \text{200}\text{°}$$ for $$\beta \in [-\text{180}\text{°};\text{180}\text{°}]$$.

$$\beta = -\text{70,5}\text{°}$$ or $$\beta = \text{109,5}\text{°}$$

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## Trig equations using CAST diagram

Subject: Mathematics

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## 3.3: Solving Trigonometric Equations

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• Page ID 61251

Learning Objectives

• Use the fundamental identities to solve trigonometric equations.
• Express trigonometric expressions in simplest form.
• Solve trigonometric equations by factoring.
• Solve trigonometric equations by using the Quadratic Formula.

Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles , which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.

In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.

## Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is $$2\pi$$. In other words, every $$2\pi$$ units, the y- values repeat. If we need to find all possible solutions, then we must add $$2\pi k$$,where $$k$$ is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is $$2\pi$$:

$\sin \theta=\sin(\theta \pm 2k\pi)$

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

Example $$\PageIndex{1A}$$: Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation $$\cos \theta=\dfrac{1}{2}$$.

From the unit circle, we know that

\begin{align*} \cos \theta &=\dfrac{1}{2} \\[4pt] \theta &=\dfrac{\pi}{3},\space \dfrac{5\pi}{3} \end{align*}

These are the solutions in the interval $$[ 0,2\pi ]$$. All possible solutions are given by

$\theta=\dfrac{\pi}{3} \pm 2k\pi \quad \text{and} \quad \theta=\dfrac{5\pi}{3} \pm 2k\pi \nonumber$

where $$k$$ is an integer.

Example $$\PageIndex{1B}$$: Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation $$\sin t=\dfrac{1}{2}$$.

Solving for all possible values of $$t$$ means that solutions include angles beyond the period of $$2\pi$$. From the section on Sum and Difference Identities, we can see that the solutions are $$t=\dfrac{\pi}{6}$$ and $$t=\dfrac{5\pi}{6}$$. But the problem is asking for all possible values that solve the equation. Therefore, the answer is

$t=\dfrac{\pi}{6}\pm 2\pi k \quad \text{and} \quad t=\dfrac{5\pi}{6}\pm 2\pi k \nonumber$

How to: Given a trigonometric equation, solve using algebra

• Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
• Substitute the trigonometric expression with a single variable, such as $$x$$ or $$u$$.
• Solve the equation the same way an algebraic equation would be solved.
• Substitute the trigonometric expression back in for the variable in the resulting expressions.
• Solve for the angle.

Example $$\PageIndex{2}$$: Solve the Linear Trigonometric Equation

Solve the equation exactly: $$2 \cos \theta−3=−5$$, $$0≤\theta<2\pi$$.

Use algebraic techniques to solve the equation.

\begin{align*} 2 \cos \theta-3&= -5\\ 2 \cos \theta&= -2\\ \cos \theta&= -1\\ \theta&= \pi \end{align*}

Exercise $$\PageIndex{1}$$

Solve exactly the following linear equation on the interval $$[0,2\pi)$$: $$2 \sin x+1=0$$.

$$x=\dfrac{7\pi}{6},\space \dfrac{11\pi}{6}$$

## Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see [link]). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is $$\pi$$,not $$2\pi$$. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of $$\dfrac{\pi}{2}$$,unless, of course, a problem places its own restrictions on the domain.

Example $$\PageIndex{3A}$$: Solving a Trignometric Equation Involving Sine

Solve the problem exactly: $$2 {\sin}^2 \theta−1=0$$, $$0≤\theta<2\pi$$.

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate $$\sin \theta$$. Then we will find the angles.

\begin{align*} 2 {\sin}^2 \theta-1&= 0\\ 2 {\sin}^2 \theta&= 1\\ {\sin}^2 \theta&= \dfrac{1}{2}\\ \sqrt{ {\sin}^2 \theta }&= \pm \sqrt{ \dfrac{1}{2} }\\ \sin \theta&= \pm \dfrac{1}{\sqrt{2}}\\ &= \pm \dfrac{\sqrt{2}}{2}\\ \theta&= \dfrac{\pi}{4}, \space \dfrac{3\pi}{4},\space \dfrac{5\pi}{4}, \space \dfrac{7\pi}{4} \end{align*}

As $$\sin \theta=−\dfrac{1}{2}$$, notice that all four solutions are in the third and fourth quadrants.

Example $$\PageIndex{3B}$$: Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: $$\csc \theta=−2$$, $$0≤\theta<4\pi$$.

We want all values of $$\theta$$ for which $$\csc \theta=−2$$ over the interval $$0≤\theta<4\pi$$.

\begin{align*} \csc \theta&= -2\\ \dfrac{1}{\sin \theta}&= -2\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\space \dfrac{11\pi}{6},\space \dfrac{19\pi}{6}, \space \dfrac{23\pi}{6} \end{align*}

Example $$\PageIndex{3C}$$: Solving an Equation Involving Tangent

Solve the equation exactly: $$\tan\left(\theta−\dfrac{\pi}{2}\right)=1$$, $$0≤\theta<2\pi$$.

Recall that the tangent function has a period of $$\pi$$. On the interval $$[ 0,\pi )$$,and at the angle of $$\dfrac{\pi}{4}$$,the tangent has a value of $$1$$. However, the angle we want is $$\left(\theta−\dfrac{\pi}{2}\right)$$. Thus, if $$\tan\left(\dfrac{\pi}{4}\right)=1$$,then

\begin{align*} \theta-\dfrac{\pi}{2}&= \dfrac{\pi}{4}\\ \theta&= \dfrac{3\pi}{4} \pm k\pi \end{align*}

Over the interval $$[ 0,2\pi )$$,we have two solutions:

$$\theta=\dfrac{3\pi}{4}$$ and $$\theta=\dfrac{3\pi}{4}+\pi=\dfrac{7\pi}{4}$$

Exercise $$\PageIndex{2}$$

Find all solutions for $$\tan x=\sqrt{3}$$.

$$\dfrac{\pi}{3}\pm \pi k$$

Example $$\PageIndex{4}$$: Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation $$2(\tan x+3)=5+\tan x$$, $$0≤x<2\pi$$.

We can solve this equation using only algebra. Isolate the expression $$\tan x$$ on the left side of the equals sign.

There are two angles on the unit circle that have a tangent value of $$−1$$: $$\theta=\dfrac{3\pi}{4}$$ and $$\theta=\dfrac{7\pi}{4}$$.

## Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

Example $$\PageIndex{5A}$$: Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation $$\sin \theta=0.8$$,where $$\theta$$ is in radians.

Make sure mode is set to radians. To find $$\theta$$, use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the $${\sin}^{−1}$$ function. What is shown on the screen is $${\sin}^{−1}$$.The calculator is ready for the input within the parentheses. For this problem, we enter $${\sin}^{−1}(0.8)$$, and press ENTER. Thus, to four decimals places,

$${\sin}^{−1}(0.8)≈0.9273$$

The solution is

$$\theta≈0.9273\pm 2\pi k$$

The angle measurement in degrees is

\begin{align*} \theta&\approx 53.1^{\circ}\\ \theta&\approx 180^{\circ}-53.1^{\circ}\\ &\approx 126.9^{\circ} \end{align*}

Note that a calculator will only return an angle in quadrants I or IV for the sine function since that is the range of the inverse sine. The other angle is obtained by using $$\pi−\theta$$.

Example $$\PageIndex{5B}$$: Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation $$\sec θ=−4,$$ giving your answer in radians.

We can begin with some algebra.

\begin{align*} \sec \theta&= -4\\ \dfrac{1}{\cos \theta}&= -4\\ \cos \theta&= -\dfrac{1}{4} \end{align*}

Check that the MODE is in radians. Now use the inverse cosine function

\begin{align*}{\cos}^{-1}\left(-\dfrac{1}{4}\right)&\approx 1.8235\\ \theta&\approx 1.8235+2\pi k \end{align*}

Since $$\dfrac{\pi}{2}≈1.57$$ and $$\pi≈3.14$$,$$1.8235$$ is between these two numbers, thus $$\theta≈1.8235$$ is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure $$\PageIndex{2}$$.

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is $$\theta '≈\pi−1.8235≈1.3181$$. The other solution in quadrant III is $$\theta '≈\pi+1.3181≈4.4597$$.

The solutions are $$\theta≈1.8235\pm 2\pi k$$ and $$\theta≈4.4597\pm 2\pi k$$.

Exercise $$\PageIndex{3}$$

Solve $$\cos \theta=−0.2$$.

$$\theta≈1.7722\pm 2\pi k$$ and $$\theta≈4.5110\pm 2\pi k$$

## Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as $$x$$ or $$u$$. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Example $$\PageIndex{6A}$$: Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: $${\cos}^2 \theta+3 \cos \theta−1=0$$, $$0≤\theta<2\pi$$.

We begin by using substitution and replacing $$\cos \theta$$ with $$x$$. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let $$\cos \theta=x$$. We have

$$x^2+3x−1=0$$

The equation cannot be factored, so we will use the quadratic formula: $$x=\dfrac{−b\pm \sqrt{b^2−4ac}}{2a}$$.

\begin{align*} x&= \dfrac{ -3\pm \sqrt{ {(-3)}^2-4 (1) (-1) } }{2}\\ &= \dfrac{-3\pm \sqrt{13}}{2}\end{align*}

Replace $$x$$ with $$\cos \theta$$ and solve.

\begin{align*} \cos \theta&= \dfrac{-3\pm \sqrt{13}}{2}\\ \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right) \end{align*}

Note that only the + sign is used. This is because we get an error when we solve $$\theta={\cos}^{−1}\left(\dfrac{−3−\sqrt{13}}{2}\right)$$ on a calculator, since the domain of the inverse cosine function is $$[ −1,1 ]$$. However, there is a second solution:

\begin{align*} \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 1.26 \end{align*}

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

\begin{align*} \theta&= 2\pi-{\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 5.02 \end{align*}

Example $$\PageIndex{6B}$$: Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: $$2 {\sin}^2 \theta−5 \sin \theta+3=0$$, $$0≤\theta≤2\pi$$.

Using grouping, this quadratic can be factored. Either make the real substitution, $$\sin \theta=u$$,or imagine it, as we factor:

\begin{align*} 2 {\sin}^2 \theta-5 \sin \theta+3&= 0\\ (2 \sin \theta-3)(\sin \theta-1)&= 0 \qquad \text {Now set each factor equal to zero.}\\ 2 \sin \theta-3&= 0\\ 2 \sin \theta&= 3\\ \sin \theta&= \dfrac{3}{2}\\ \sin \theta-1&= 0\\ \sin \theta&= 1 \end{align*}

Next solve for $$\theta$$: $$\sin \theta≠\dfrac{3}{2}$$, as the range of the sine function is $$[ −1,1 ]$$. However, $$\sin \theta=1$$, giving the solution $$\theta=\dfrac{\pi}{2}$$.

Make sure to check all solutions on the given domain as some factors have no solution.

Exercise $$\PageIndex{4}$$

Solve $${\sin}^2 \theta=2 \cos \theta+2$$, $$0≤\theta≤2\pi$$. [Hint: Make a substitution to express the equation only in terms of cosine.]

$$\cos \theta=−1$$, $$\theta=\pi$$

Example $$\PageIndex{7A}$$: Solving a Trigonometric Equation Using Algebra

Solve exactly: $$2 {\sin}^2 \theta+\sin \theta=0;\space 0≤\theta<2\pi$$

This problem should appear familiar as it is similar to a quadratic. Let $$\sin \theta=x$$. The equation becomes $$2x^2+x=0$$. We begin by factoring:

\begin{align*} 2x^2+x&= 0\\ x(2x+1)&= 0\qquad \text {Set each factor equal to zero.}\\ x&= 0\\ 2x+1&= 0\\ x&= -\dfrac{1}{2} \end{align*} Then, substitute back into the equation the original expression $$\sin \theta$$ for $$x$$. Thus, \begin{align*} \sin \theta&= 0\\ \theta&= 0,\pi\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6} \end{align*}

The solutions within the domain $$0≤\theta<2\pi$$ are $$\theta=0,\pi,\dfrac{7\pi}{6},\dfrac{11\pi}{6}$$.

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

\begin{align*} {\sin}^2 \theta+\sin \theta&= 0\\ \sin \theta(2\sin \theta+1)&= 0\\ \sin \theta&= 0\\ \theta&= 0,\pi\\ 2 \sin \theta+1&= 0\\ 2\sin \theta&= -1\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6} \end{align*}

We can see the solutions on the graph in Figure $$\PageIndex{3}$$. On the interval $$0≤\theta<2\pi$$,the graph crosses the $$x$$ - axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

We can verify the solutions on the unit circle in via the result in the section on Sum and Difference Identities as well.

Example $$\PageIndex{7B}$$: Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: $$2 {\sin}^2 \theta−3 \sin \theta+1=0$$, $$0≤\theta<2\pi$$.

We can factor using grouping. Solution values of $$\theta$$ can be found on the unit circle.

\begin{align*} (2 \sin \theta-1)(\sin \theta-1)&= 0\\ 2 \sin \theta-1&= 0\\ \sin \theta&= \dfrac{1}{2}\\ \theta&= \dfrac{\pi}{6}, \dfrac{5\pi}{6}\\ \sin \theta&= 1\\ \theta&= \dfrac{\pi}{2} \end{align*}

Exercise $$\PageIndex{5}$$

Solve the quadratic equation $$2{\cos}^2 \theta+\cos \theta=0$$.

$$\dfrac{\pi}{2}, \space \dfrac{2\pi}{3}, \space \dfrac{4\pi}{3}, \space \dfrac{3\pi}{2}$$

## Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

Example $$\PageIndex{8}$$: Solving an Equation Using an Identity

Solve the equation exactly using an identity: $$3 \cos \theta+3=2 {\sin}^2 \theta$$, $$0≤\theta<2\pi$$.

If we rewrite the right side, we can write the equation in terms of cosine:

\begin{align*} 3 \cos \theta+3&= 2 {\sin}^2 \theta\\ 3 \cos \theta+3&= 2(1-{\cos}^2 \theta)\\ 3 \cos \theta+3&= 2-2{\cos}^2 \theta\\ 2 {\cos}^2 \theta+3 \cos \theta+1&= 0\\ (2 \cos \theta+1)(\cos \theta+1)&= 0\\ 2 \cos \theta+1&= 0\\ \cos \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{2\pi}{3},\space \dfrac{4\pi}{3}\\ \cos \theta+1&= 0\\ \cos \theta&= -1\\ \theta&= \pi\\ \end{align*}

Our solutions are $$\theta=\dfrac{2\pi}{3},\space \dfrac{4\pi}{3},\space \pi$$.

## Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as $$\sin(2x)$$ or $$\cos(3x)$$. When confronted with these equations, recall that $$y=\sin(2x)$$ is a horizontal compression by a factor of 2 of the function $$y=\sin x$$. On an interval of $$2\pi$$,we can graph two periods of $$y=\sin(2x)$$,as opposed to one cycle of $$y=\sin x$$. This compression of the graph leads us to believe there may be twice as many x -intercepts or solutions to $$\sin(2x)=0$$ compared to $$\sin x=0$$. This information will help us solve the equation.

Example $$\PageIndex{9}$$: Solving a Multiple Angle Trigonometric Equation

Solve exactly: $$\cos(2x)=\dfrac{1}{2}$$ on $$[ 0,2\pi )$$.

We can see that this equation is the standard equation with a multiple of an angle. If $$\cos(\alpha)=\dfrac{1}{2}$$,we know $$\alpha$$ is in quadrants I and IV. While $$\theta={\cos}^{−1} \dfrac{1}{2}$$ will only yield solutions in quadrants I and II, we recognize that the solutions to the equation $$\cos \theta=\dfrac{1}{2}$$ will be in quadrants I and IV.

Therefore, the possible angles are $$\theta=\dfrac{\pi}{3}$$ and $$\theta=\dfrac{5\pi}{3}$$. So, $$2x=\dfrac{\pi}{3}$$ or $$2x=\dfrac{5\pi}{3}$$, which means that $$x=\dfrac{\pi}{6}$$ or $$x=\dfrac{5\pi}{6}$$. Does this make sense? Yes, because $$\cos\left(2\left(\dfrac{\pi}{6}\right)\right)=\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}$$.

In quadrant I, $$2x=\dfrac{\pi}{3}$$, so $$x=\dfrac{\pi}{6}$$ as noted. Let us revolve around the circle again:

\begin{align*} 2x&= \dfrac{\pi}{3}+2\pi\\ &= \dfrac{\pi}{3}+\dfrac{6\pi}{3}\\ &= \dfrac{7\pi}{3}\\ x&= \dfrac{7\pi}{6}\\ \text {One more rotation yields}\\ 2x&= \dfrac{\pi}{3}+4\pi\\ &= \dfrac{\pi}{3}+\dfrac{12\pi}{3}\\ &= \dfrac{13\pi}{3}\\ \end{align*}

$$x=\dfrac{13\pi}{6}>2\pi$$, so this value for $$x$$ is larger than $$2\pi$$, so it is not a solution on $$[ 0,2\pi )$$.

In quadrant IV, $$2x=\dfrac{5\pi}{3}$$, so $$x=\dfrac{5\pi}{6}$$ as noted. Let us revolve around the circle again:

\begin{align*} 2x&= \dfrac{5\pi}{3}+2\pi\\ &= \dfrac{5\pi}{3}+\dfrac{6\pi}{3}\\ &= \dfrac{11\pi}{3} \end{align*}

so $$x=\dfrac{11\pi}{6}$$.

One more rotation yields

\begin{align*} 2x&= \dfrac{5\pi}{3}+4\pi\\ &= \dfrac{5\pi}{3}+\dfrac{12\pi}{3}\\ &= \dfrac{17\pi}{3} \end{align*}

$$x=\dfrac{17\pi}{6}>2\pi$$,so this value for $$x$$ is larger than $$2\pi$$,so it is not a solution on $$[ 0,2\pi )$$ ﻿ .

Our solutions are $$x=\dfrac{\pi}{6}, \space \dfrac{5\pi}{6}, \space \dfrac{7\pi}{6}$$, and $$\dfrac{11\pi}{6}$$. Note that whenever we solve a problem in the form of $$sin(nx)=c$$, we must go around the unit circle $$n$$ times.

## Key Concepts

• When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. See Example $$\PageIndex{1}$$, Example $$\PageIndex{2}$$, and Example $$\PageIndex{3}$$.
• Equations involving a single trigonometric function can be solved or verified using the unit circle. See Example $$\PageIndex{4}$$, Example $$\PageIndex{5}$$, and Example $$\PageIndex{6}$$, and Example $$\PageIndex{7}$$.
• We can also solve trigonometric equations using a graphing calculator. See Example $$\PageIndex{8}$$ and Example $$\PageIndex{9}$$.
• Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See Example $$\PageIndex{10}$$, Example $$\PageIndex{11}$$, Example $$\PageIndex{12}$$, and Example $$\PageIndex{13}$$.
• We can also use the identities to solve trigonometric equation. See Example $$\PageIndex{14}$$, Example $$\PageIndex{15}$$, and Example $$\PageIndex{16}$$.
• We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See Example $$\PageIndex{17}$$.
• Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See Example $$\PageIndex{18}$$.

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## Course content

• Understand how sin, cos and tan are defined for angles from 0° to 360°
• Understand period and related angles
• Solve basic trigonometric equations.

## Related angles

The trig functions sin, cos and tan can be defined for angles greater than 90° by thinking about this diagram:

The four quadrants contain congruent right-angled triangles with the angle $$x^\circ$$ at the centre.

The actual angle is measured anticlockwise from the 0° line .

• 1st quadrant: The actual angle is $$x^\circ.$$
• 2nd quadrant: The actual angle is half a turn and then back $$x^\circ.$$ So that's 180–$$x^\circ.$$
• 3rd quadrant: The actual angle is half a turn and then another $$x^\circ$$. So that's 180+$$x^\circ.$$
• 4th quadrant: The actual angle is a full turn and then back $$x^\circ.$$ So that's 360–$$x^\circ.$$

These four angles are what we call related .

Example 1:   $$x$$ = 30°

• sin 30° = $$1 \over 2$$
• sin (180–30)° = sin 150° = $$1 \over 2$$
• sin (180+30)° = sin 210° = $$-\frac{1}{2}$$
• sin (360–30)° = sin 330° = $$-\frac{1}{2}$$

Apart from the negatives, they're all equal. That's because these four angles are related.

Example 2:   $$x$$ = 45°

• cos 45° = $$\frac{\sqrt{2}}{2}$$
• cos (180–45)° = cos 135° = $$-\frac{\sqrt{2}}{2}$$
• cos (180+45)° = cos 225° = $$-\frac{\sqrt{2}}{2}$$
• cos (360–45)° = cos 315° = $$\frac{\sqrt{2}}{2}$$

Again, these are related angles, so they're equal apart from the negatives.

Example 3:   $$x$$ = 60°

• tan 60° = $$\sqrt{3}$$
• tan (180–60)° = tan 120° = $$-\sqrt{3}$$
• tan (180+60)° = tan 240° = $$\sqrt{3}$$
• tan (360–60)° = tan 300° = $$-\sqrt{3}$$

Same again: they're equal apart from the negatives, because they are related angles.

## ASTC diagram

This is also called a CAST diagram, but we prefer ASTC as it keeps the letters in the correct quadrant order.

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Here is a simpler version of the diagram:

If you want to understand why some of the values for related angles are positive and some are negative, that's easy. Just go back to your study of the basic trig graphs and take a look at when they are above or below the $$x$$-axis. Above is positive and below is negative – simple!

## Trig equations

Trig equations have an infinity of solutions because the wave-like graphs of these functions repeat endlessly.

For this reason, Nat 5 exams ask for such equations to be solved within a given domain , often $$0 \leq x \lt 360.$$

When solving any trig equation, we recommend always finding the related acute angle . This means the 1st quadrant (0° to 90°) angle that is related to the solutions.

We will use the Greek letter alpha ($$\alpha$$) for the related acute angle.

Once we know the related acute angle, we will use it to find the solutions.

## Find a Maths tutor

Solve the equation $$cos\ x^\circ=0.5$$, for $$0 \leq x \lt 360.$$

The related acute angle, $$\alpha\ = cos^{-1}\ 0.5=60^\circ$$

Because $$cos\ x$$ > 0, there are solutions in the 1 st (A) and 4 th (C) quadrants.

So $$x=\alpha$$  or  $$x=360-\alpha$$

$$x=60^\circ$$  or  $$x=360-60=300^\circ$$

Solve the equation $$cos\ x^\circ=-0.5$$, for $$0 \leq x \lt 360.$$

The only difference here is the negative.

The related acute angle is the same: $$\alpha\ = cos^{-1}\ 0.5=60^\circ$$

Because $$cos\ x$$ < 0, there are solutions in the 2 nd (S) and 3 rd (T) quadrants.

So $$x=180-\alpha$$  or  $$x=180+\alpha$$

$$x=180-60=120^\circ$$  or  $$x=180+60=240^\circ$$

Solve the equation $$sin\ x^\circ=-0.4$$, for $$0 \leq x \lt 360.$$

The related acute angle $$\alpha\ = sin^{-1}\ 0.4 = 23.6^\circ$$ (to 1 d.p.)

Because $$sin\ x$$ < 0, there are solutions in the 3 rd (T) and 4 th (C) quadrants.

So $$x=180+\alpha$$  or  $$x=360-\alpha$$

$$x=180+23.6=203.6^\circ$$  or  $$x=360-23.6=336.4^\circ$$ (to 1 d.p.)

Solve the equation $$tan\ x^\circ=5$$, for $$0 \leq x \lt 360.$$

The related acute angle $$\alpha\ = tan^{-1}\ 5 = 78.7^\circ$$ (to 1 d.p.)

Because $$tan\ x$$ > 0, there are solutions in the 1 st (A) and 3 rd (T) quadrants.

So $$x=\alpha$$  or  $$x=180+\alpha$$

$$x=78.7^\circ$$  or  $$x=180+78.7=258.7^\circ$$ (to 1 d.p.)

Solve the equation $$9\tiny\ \normalsize tan\ x^\circ-1=6$$, for $$0 \leq x \lt 360.$$

The previous examples were easier than exam level. This is more like it!

First we rearrange to find a value for $$tan\ x$$:

$$9\tiny\ \normalsize tan\ x^\circ-1=6$$ $$9\tiny\ \normalsize tan\ x^\circ=7$$ $$tan\ x^\circ=\large \frac{7}{9} \normalsize$$

The related acute angle $$\alpha\ = tan^{-1}\ \large \frac{7}{9} \normalsize = 37.9^\circ$$ (to 1 d.p.)

$$x=37.9^\circ$$  or  $$x=180+37.9=217.9^\circ$$ (to 1 d.p.)

Solve the equation $$3\tiny\ \normalsize sin\ x^\circ+5=4$$, for $$0 \leq x \lt 360.$$

Rearrange to make $$sin\ x$$ the subject:

$$3\tiny\ \normalsize sin\ x^\circ+5=4$$ $$3\tiny\ \normalsize sin\ x^\circ=-1$$ $$sin\ x^\circ=-\large \frac{1}{3} \normalsize$$

The related acute angle $$\alpha\ = sin^{-1}\ \large \frac{1}{3} \normalsize = 19.5^\circ$$ (to 1 d.p.)

$$x=180+19.5=199.5^\circ$$  or  $$x=360-19.5=340.5^\circ$$ (to 1 d.p.)

## Example 7 (non-calculator)

Write the following in order of size, starting with the smallest.

$$sin\ 200^\circ$$   $$sin\ 160^\circ$$   $$sin\ 180^\circ$$

Obviously this isn't an equation, but a question very like it appeared as Q9 on 2015 Paper 1 .

Answering it without a calculator requires either sketching the graph of $$y=sin\ x$$ or thinking about the ASTC diagram.

• $$200^\circ$$ is in the 3 rd (T) quadrant, so $$sin\ 200^\circ \lt 0$$
• $$160^\circ$$ is in the 2 nd (S) quadrant, so $$sin\ 160^\circ \gt 0$$
• $$180^\circ$$ is between the 2 nd and 3 rd quadrants, so $$sin\ 180^\circ=0$$

In order:  $$sin\ 200^\circ$$  $$sin\ 180^\circ$$  $$sin\ 160^\circ$$

SQA National 5 Maths 2014 P2 Q12

Solve the equation $$11\tiny\ \normalsize cos\ x^\circ-2=3$$, for $$0 \leq x \leq 360.$$

$$11\tiny\ \normalsize cos\ x^\circ-2=3$$ $$11\tiny\ \normalsize cos\ x^\circ=5$$ $$cos\ x^\circ=\large \frac{5}{11} \normalsize$$

The related acute angle $$\alpha\ = cos^{-1}\ \large \frac{5}{11} \normalsize = 63.0^\circ$$ (to 1 d.p.)

$$x=63.0^\circ$$  or  $$x=360-63.0=297.0^\circ$$ (to 1 d.p.)

SQA National 5 Maths 2016 P2 Q14

Solve the equation $$2\tiny\ \normalsize tan\ x^\circ+5=-4$$, for $$0 \leq x \leq 360.$$

$$2\tiny\ \normalsize tan\ x^\circ+5=-4$$ $$2\tiny\ \normalsize tan\ x^\circ=-9$$ $$tan\ x^\circ=-\large \frac{9}{2} \normalsize$$

The related acute angle $$\alpha\ = tan^{-1}\ \large \frac{9}{2} \normalsize = 77.5^\circ$$ (to 1 d.p.)

Because $$tan\ x$$ < 0, there are solutions in the 2 nd (S) and 4 th (C) quadrants.

So $$x=180-\alpha$$  or  $$x=360-\alpha$$

$$x=180-77.5=102.5^\circ$$  or  $$x=360-77.5=282.5^\circ$$ (to 1 d.p.)

SQA National 5 Maths 2018 P2 Q8

Solve the equation $$7\tiny\ \normalsize sin\ x^\circ+2=3$$, for $$0 \leq x \lt 360.$$

$$7\tiny\ \normalsize sin\ x^\circ+2=3$$ $$7\tiny\ \normalsize sin\ x^\circ=1$$ $$sin\ x^\circ=\large \frac{1}{7} \normalsize$$

The related acute angle $$\alpha\ = sin^{-1}\ \large \frac{1}{7} \normalsize = 8.2^\circ$$ (to 1 d.p.)

Because $$sin\ x$$ > 0, there are solutions in the 1 st (A) and 2 nd (S) quadrants.

So $$x=\alpha$$  or  $$x=180-\alpha$$

$$x=8.2^\circ$$  or  $$x=180-8.2=171.8^\circ$$ (to 1 d.p.)

SQA National 5 Maths 2019 P2 Q14

Solve the equation $$5\tiny\ \normalsize cos\ x^\circ+2=1$$, for $$0 \leq x \leq 360.$$

$$5\tiny\ \normalsize cos\ x^\circ+2=1$$ $$5\tiny\ \normalsize cos\ x^\circ=-1$$ $$cos\ x^\circ=-\large \frac{1}{5} \normalsize$$

The related acute angle $$\alpha\ = cos^{-1}\ \large \frac{1}{5} \normalsize = 78.5^\circ$$ (to 1 d.p.)

$$x=180-78.5=101.5^\circ$$  or  $$x=180+78.5=258.5^\circ$$ (to 1 d.p.)

SQA National 5 Maths 2022 P2 Q9

Solve the equation $$3\tiny\ \normalsize sin\ x^\circ+4=6$$, for $$0 \leq x \leq 360.$$

$$3\tiny\ \normalsize sin\ x^\circ+4=6$$ $$3\tiny\ \normalsize sin\ x^\circ=2$$ $$sin\ x^\circ=\large \frac{2}{3} \normalsize$$

The related acute angle $$\alpha\ = sin^{-1}\ \large \frac{2}{3} \normalsize = 41.8^\circ$$ (to 1 d.p.)

$$x=41.8^\circ$$  or  $$x=180-41.8=138.2^\circ$$ (to 1 d.p.)

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## Introduction to Trigonometry

Trigonometry (from Greek trigonon "triangle" + metron "measure")

Want to learn Trigonometry? Here is a quick summary. Follow the links for more, or go to Trigonometry Index

Trigonometry helps us find angles and distances, and is used a lot in science, engineering, video games, and more!

## Right-Angled Triangle

The triangle of most interest is the right-angled triangle . The right angle is shown by the little box in the corner:

Another angle is often labeled θ , and the three sides are then called:

• Opposite : opposite the angle θ
• and the longest side is the Hypotenuse

## Why a Right-Angled Triangle?

Why is this triangle so important?

Imagine we can measure along and up but want to know the direct distance and angle:

Trigonometry can find that missing angle and distance.

Or maybe we have a distance and angle and need to "plot the dot" along and up:

Questions like these are common in engineering, computer animation and more.

## Sine, Cosine and Tangent

The main functions in trigonometry are Sine, Cosine and Tangent

They are simply one side of a right-angled triangle divided by another.

For any angle " θ ":

(Sine, Cosine and Tangent are often abbreviated to sin, cos and tan .)

## Example: What is the sine of 35°?

Using this triangle (lengths are only to one decimal place):

sin(35°) = Opposite Hypotenuse = 2.8 4.9 = 0.57...

The triangle could be larger, smaller or turned around, but that angle will always have that ratio .

Calculators have sin, cos and tan to help us, so let's see how to use them:

## Example: How Tall is The Tree?

We can't reach the top of the tree, so we walk away and measure an angle (using a protractor) and distance (using a laser):

• We know the Hypotenuse
• And we want to know the Opposite

Sine is the ratio of Opposite / Hypotenuse :

sin(45°) = Opposite Hypotenuse

Get a calculator, type in "45", then the "sin" key:

sin(45°) = 0.7071...

What does the 0.7071... mean? It is the ratio of the side lengths, so the Opposite is about 0.7071 times as long as the Hypotenuse.

We can now put 0.7071... in place of sin(45°):

0.7071... = Opposite Hypotenuse

And we also know the hypotenuse is 20 :

0.7071... = Opposite 20

To solve, first multiply both sides by 20:

20 × 0.7071... = Opposite

Opposite = 14.14m (to 2 decimals)

The tree is 14.14m tall

## Try Sin Cos and Tan

Play with this for a while (move the mouse around) and get familiar with values of sine, cosine and tangent for different angles, such as 0°, 30°, 45°, 60° and 90°.

Also try 120°, 135°, 180°, 240°, 270° etc, and notice that positions can be positive or negative by the rules of Cartesian coordinates , so the sine, cosine and tangent change between positive and negative also.

So trigonometry is also about circles !

## Unit Circle

What you just played with is the Unit Circle .

It is a circle with a radius of 1 with its center at 0.

Because the radius is 1, we can directly measure sine, cosine and tangent.

Here we see the sine function being made by the unit circle:

Note: you can see the nice graphs made by sine, cosine and tangent .

Angles can be in Degrees or Radians . Here are some examples:

## Repeating Pattern

Because the angle is rotating around and around the circle the Sine, Cosine and Tangent functions repeat once every full rotation (see Amplitude, Period, Phase Shift and Frequency ).

When we want to calculate the function for an angle larger than a full rotation of 360° (2 π radians) we subtract as many full rotations as needed to bring it back below 360° (2 π radians):

## Example: what is the cosine of 370°?

370° is greater than 360° so let us subtract 360°

370° − 360° = 10°

cos(370°) = cos(10°) = 0.985 (to 3 decimal places)

And when the angle is less than zero, just add full rotations.

## Example: what is the sine of −3 radians?

−3 is less than 0 so let us add 2 π radians

−3 + 2 π = −3 + 6.283... = 3.283... rad ians

sin(−3) = sin(3.283...) = −0.141 (to 3 decimal places)

## Solving Triangles

Trigonometry is also useful for general triangles, not just right-angled ones .

It helps us in Solving Triangles . "Solving" means finding missing sides and angles.

## Example: Find the Missing Angle "C"

Angle C can be found using angles of a triangle add to 180° :

So C = 180° − 76° − 34° = 70°

We can also find missing side lengths. The general rule is:

When we know any 3 of the sides or angles we can find the other 3 (except for the three angles case)

See Solving Triangles for more details.

## Other Functions (Cotangent, Secant, Cosecant)

Similar to Sine, Cosine and Tangent, there are three other trigonometric functions which are made by dividing one side by another:

## Trigonometric and Triangle Identities

And as you get better at Trigonometry you can learn these:

Enjoy becoming a triangle (and circle) expert!

#### IMAGES

1. Mathematics Trigonometry basics-CAST diagram for grades 11 & 12 and N3

2. Trigonometry CAST Diagrams and Solving Equations

3. Trig. Equations Examples using CAST Diagrams (solutions, examples

4. Solving trig equations using a CAST diagram

5. solving trig equations using cast diagram

6. How to use CAST method for solving trig equations

1. Trig. Equations Examples using CAST Diagrams

The Cast diagram helps us to remember the signs of the trigonometric functions in each of the quadrants. The CAST diagram is also called the Quadrant Rule or the ASTC diagram. In the first quadrant, the values are all positive. In the second quadrant, only the values for sin are positive. In the third quadrant, only the values for tan are positive.

2. An Introduction to Solving Trigonometric Equations

This video explains how the CAST diagram can be used to solve trig equations. Solving Trig Equations 1: • Solving Trigonometric Equations 1 Practice Questions:...

3. Solving Trigonometric Equations 1

This video shows how to solve trig equations using the CAST diagram. Introduction: https://youtu.be/kJwCXuxLj4EPractice Questions: https://corbettmaths.com/w...

4. Solving trigonometric equations in degrees

Solving trigonometric equations Solving trigonometric equations in degrees. Trigonometric equations can be solved in degrees or radians using CAST and its period to find other solutions within the ...

5. Using the CAST method to solve trigonometric equations

0:00 / 13:05 Using the CAST method to solve trigonometric equations. Maths-Boost 553 subscribers Subscribe Subscribed 54 4.3K views 5 years ago A Level Pure A full explanation of how to use...

6. PDF Trigonometric equations LESSON

When solving trigonometric equations, the CAST diagram can be used to help calculate all the required angles. There are two important points to remember about using the CAST diagram: All angles are drawn from the horizontal. All angles are measured anticlockwise from the positive x− axis. Solving equations using the CAST diagram

7. Introduction to Trig Equations Video

A video introducing how to solve trig equations using the CAST diagram. Welcome; Videos and Worksheets; Primary; 5-a-day. 5-a-day GCSE 9-1; 5-a-day Primary; 5-a-day Further Maths; More. Further Maths; GCSE Revision; Revision Cards; Books; Introduction to Trig Equations Video . Videos. Previous: October Update. Next: Solving Trig Equations 1 ...

8. Solving trig equations using a CAST diagram

Solving trig equations using a CAST diagram Subject: Mathematics Age range: 16+ Resource type: Other File previews pptx, 338.16 KB Recaps where a CAST diagram comes from, relating to the graphs. Then moves onto solving more complex equations using the diagram as this was not the first lesson on solving trig equations. Creative Commons "Sharealike"

9. Trigonometric equations in radians (extension)

Trigonometric equations can be solved in degrees or radians using CAST and its period to find other solutions within the range, including multiple or compound angles and the wave function....

10. 6.4 Trigonometric equations

Solving trigonometric equations requires that we find the value of the angles that satisfy the equation. If a specific interval for the solution is given, then we need only find the value of the angles within the given interval that satisfy the equation. ... Use the CAST diagram to determine in which quadrants $$\tan \theta$$ is positive. The ...

11. Trigonometry CAST Diagrams and Solving Equations

Trigonometry at AS/A Level includes solving trigonometric equations, and finding multiple solutions using a CAST diagram (or graph). In this video I introduce the CAST diagram and...

12. Solving trigonometric equations CAST

10 February 2021 Not quite what you were looking for? Search by keyword to find the right resource: Presentation with examples on how to solve trigonometric equations using the CAST diagram. AS Maths

13. Trig equations using CAST diagram

Trig equations using CAST diagram Subject: Mathematics Age range: 5-7 Resource type: Worksheet/Activity File previews docx, 177.47 KB pdf, 398.55 KB I'm a little bit torn when it comes to teaching trig equations using the cast diagram. It seems like teaching without the understanding, however I find they get the right answer more often with it.

14. 3.3: Solving Trigonometric Equations

Solving Trigonometric Equations with Multiple Angles. Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as $$\sin(2x)$$ or $$\cos(3x)$$. When confronted with these equations, recall that $$y=\sin(2x)$$ is a horizontal compression by a factor of 2 of the function $$y=\sin x$$.

15. The CAST Diagram

The CASt diagram provides a method to find multiple solutions to trigonometric equations. It indicates the quadrants in which the elementary trigonometric functions are positive (or negative). It is normally shown as below. The significance of each eletter is shown in the diagram below. Example: To solve in the range 0° - 360°, find Since ...

16. How to use CAST diagrams to solve Trigonometric Equations

In this lesson you will learn how to use the CAST diagram to solve Trigonometric Equations for Year 1 A Level Maths.Visit our new website www.mathscoachacade...

17. trigonometry

Using cast diagram to solve equation Asked 6 years, 10 months ago Modified 6 years, 10 months ago Viewed 939 times 0 I have the question Solve each equation for X X in the interval 0 ≤ X ≤ 360 0 ≤ X ≤ 360 sin(X − 30) = −0.3 sin ( X − 30) = − 0.3 Here is my attempt: sin sin is negative and so tan tan and cos cos are positive.

18. 3.4.2 Linear Trigonometric Equations

Another way to find solutions is by using the CAST diagram which shows where each function has positive solutions You may be asked to use degrees or radians to solve trigonometric equations Make sure your calculator is in the correct mode Remember common angles 90° is ½π radians 180° is π radians 270° is 3π/2 radians 360° is 2π radians

19. Trig Equations

Trig equations. Trig equations have an infinity of solutions because the wave-like graphs of these functions repeat endlessly. For this reason, Nat 5 exams ask for such equations to be solved within a given domain, often $$0 \leq x \lt 360.$$ When solving any trig equation, we recommend always finding the related acute angle.

20. How to solve trigonometric equations using the Quadrant Rule

Here I show you how to use the quadrant rule or CAST diagram to solve trigonometric equations in different ranges. It saves having to use trigonometric graph...

21. Trigonometry

Other Functions (Cotangent, Secant, Cosecant) Similar to Sine, Cosine and Tangent, there are three other trigonometric functions which are made by dividing one side by another: Cosecant Function: csc (θ) = Hypotenuse / Opposite. Secant Function: sec (θ) = Hypotenuse / Adjacent. Cotangent Function: cot (θ) = Adjacent / Opposite.

22. Using Trig Identities and CAST Diagrams to solve Trigonometric

73 Maths Coach In this second lesson on CAST diagrams, you will learn how to combine the trigonometric identities and cast diagrams to solve questions for Year 1 A Level Ma...