Topics
6. Trigonometric Functions
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United States of America
Grades 9-12

6.03 Solving trigonometric equations

Lesson
Practice

Introduction

As we saw in Math 2 lesson  10.04 Solving right triangles  , we can use the inverse functions to solve one step trigonometric equations like \text{sin }\theta = \dfrac{1}{2}, by taking the applying the inverse function to both sides of the equation to give \theta = \text{sin}^{-1}\left(\dfrac{1}{2}\right).

We will now take it one step further and require some algebraic manipulation before taking the trigonometric inverse.

Solve using reciprocal functions

To solve an equation involving reciprocal trigonometric functions, recall their relationship to the trigonometric functions:

\displaystyle \text{csc }\theta\displaystyle =\displaystyle \dfrac{1}{\text{sin }\theta}
\displaystyle \text{sec }\theta\displaystyle =\displaystyle \dfrac{1}{\text{cos }\theta}
\displaystyle \text{cot }\theta\displaystyle =\displaystyle \dfrac{\text{cos }\theta}{\text{sin }\theta} = \dfrac{1}{\text{tan }\theta}

Examples

Example 1

Find the measure in degrees of the acute angle satisfying \text{tan }\theta = 5\sqrt{3}-4 \text{ tan }\theta.

Worked Solution
Create a strategy

First we will isolate \text{ tan } \theta on one side of the equation, leaving the constant term on the other side.

Apply the idea
\displaystyle 5\text{tan } \theta\displaystyle =\displaystyle 5\sqrt{3}Add 4\text{ tan } \theta to both sides
\displaystyle \text{tan } \theta\displaystyle =\displaystyle \sqrt{3}Divide both sides by 5
\displaystyle \theta\displaystyle =\displaystyle \tan^{-1}\sqrt{3}Take the inverse of \tan on both sides
\displaystyle =\displaystyle 60\degreeEvaluate

Example 2

Find the measure in degrees of the angle satisfying 8 \text{ cos }\theta - 4 = 0 for 0\degree \lt \theta \lt 90\degree.

Worked Solution
Create a strategy

First we will isolate \text{ cos } \theta on one side of the equation, leaving the constant term on the other side.

Apply the idea
\displaystyle 8\cos \theta \displaystyle =\displaystyle 4Add 4 to both sides
\displaystyle \cos \theta\displaystyle =\displaystyle \dfrac{1}{2}Divide both sides by 8
\displaystyle \theta\displaystyle =\displaystyle \cos^{-1}\dfrac{1}{2}Take the inverse of \cos on both sides
\displaystyle =\displaystyle 60\degreeEvaluate
Idea summary

Remember the relationship between the reciprocal trigonometric functions and the trigonometric functions:

\displaystyle \text{cosec }\theta\displaystyle =\displaystyle \dfrac{1}{\text{sin }\theta}
\displaystyle \text{sec }\theta\displaystyle =\displaystyle \dfrac{1}{\text{cos }\theta}
\displaystyle \text{cot }\theta\displaystyle =\displaystyle \dfrac{\text{cos }\theta}{\text{sin }\theta} = \dfrac{1}{\text{tan }\theta}

Solve using the unit circle

Trigonometric equations express the value of a trigonometric function, when we do not know what angle leads to that value. For example, we may know that for some angle \theta we have \sin \theta = \dfrac{\sqrt{3}}{2} or \cos \beta = - \dfrac{1}{2}. The problem is to determine what values of \theta and \beta will make these true statements.

The image shows coordinate plane with a cicle and has a right triangle. Ask your teacher for more information.

As we saw in lesson  6.02 Evaluating trigonometric functions  , we can use the fact that for any point on the unit circle:

x=\cos \theta

y=\sin \theta

\dfrac{y}{x}=\tan \theta

Using the blue unit circle below, there are two instances where the \sin function is equal to \\\dfrac{\sqrt{3}}{2} (60\degree and 120\degree) and two where \cos is equal to -\dfrac{1}{2} (120\degree and 240\degree).

The image shows 2 different circles with different measurements. Ask your teacher for more information.

Examples

Example 3

Solve \sin \theta = 0 for \theta over the domain 0\degree \leq \theta \leq 360\degree.

The image shows curves that crosses in the x-axis. Ask your teacher for more information.
Worked Solution
Create a strategy

Find where y=0. That is, where the curve crosses the x-axis.

Apply the idea

The curve crosses the x-axis three times, so the solutions are:

\displaystyle \theta\displaystyle =\displaystyle 0\degree,\,180\degree,\,360\degree
Idea summary

If we have a unit circle diagram, we can use the fact that for any point on the unit circle:

\displaystyle x\displaystyle =\displaystyle \cos \theta
\displaystyle y\displaystyle =\displaystyle \sin \theta
\displaystyle \dfrac{y}{x}\displaystyle =\displaystyle \tan \theta

Solve further trigonometric equations

Given a particular value of a trigonometric function, we may wish to find the values of the domain variable that are mapped to that function value. Or, given two distinct functions defined on the same domain, we may ask what values of the domain variable make the two functions equal.

These situations are what is meant by the idea of solving trigonometric (and other) equations. We are finding the values of the variable that make the equations true.

Examples

Example 4

Given the function defined by 2\sin x = 1 for x \in [0,\,360\degree], find the values of x.

Worked Solution
Create a strategy

We can use algebra to isolate \text{ sin } x. Then, use the inverse of the \sin function to find the solution between 0\degree and 90\degree. To find the remaining solutions, we will use \sin(180\degree - x).

Apply the idea
\displaystyle 2\sin x\displaystyle =\displaystyle 1Original equation
\displaystyle \sin x\displaystyle =\displaystyle \dfrac{1}{2}Divide both sides by 2
\displaystyle x\displaystyle =\displaystyle \sin^{-1}\left(\dfrac{1}{2}\right)Take the inverse of \sin on both sides
\displaystyle =\displaystyle 30\degreeEvaluate

To find additional solutions between 0\degree and 360\degree, substitute x=30 \degree into the \sin(180\degree - x):

\sin (180\degree - 30\degree) = \sin 150\degree

We can give the solution as x=30\degree,\,150\degree

Reflect and check

Another way to identify the solutions is to think of 2\sin x=1 as a system of the equations {y=2\sin x} and y=1. To solve 2\sin x = 1, is the same as graphically finding the intersection of these two separate curves.

The image shows curves that crosses in the x-axis. It has a straight line that crosses the upper curve. Ask your teacher for more information.

Example 5

Solve 4\cos x + 1 = 3 for x \in [0,\,360\degree].

Worked Solution
Create a strategy

We can use algebra to isolate \text{ cos } x. Then, use the inverse of the \cos function to find one solution. To find the remaining solutions, we will use \cos(360\degree - x).

Apply the idea
\displaystyle 4\cos x + 1\displaystyle =\displaystyle 3Original equation
\displaystyle 4\cos x\displaystyle =\displaystyle 2Subtract 1 from both sides
\displaystyle \cos x\displaystyle =\displaystyle \dfrac{1}{2}Divide both sides by 4
\displaystyle x\displaystyle =\displaystyle \cos^{-1}\left(\dfrac{1}{2}\right)Take the inverse of \cos on both sides
\displaystyle =\displaystyle 60\degreeEvaluate

To find all solutions between 0\degree and 360\degree, substitute x=60\degree into \cos{360\degree-x}:

\cos (360\degree - 60\degree) = \cos 300\degree

We can give the solution as x=60\degree,\,300\degree

Reflect and check

The graphs for this can also be used to find the solution. Graph both y=4\cos x + 1 and y=3 on the same set of axes. The intersections of these curves are the solutions.

The image shows curves that crosses in the x-axis. It has a straight line that crosses the upper curve. Ask your teacher for more information.

Trigonometric equations do not always have easy solutions. It can be necessary to use a numerical method to obtain an approximate solution.

Example 6

Find all the solutions of the equation \sin t = 2\sin t - \dfrac{1}{2}in the interval [0,\,2\pi].

Worked Solution
Create a strategy

Use algebra to isolate \sin {x} and then review the unit circle to determine if the solution will be an exact value or an approximation.

Apply the idea
\displaystyle \sin t\displaystyle =\displaystyle 2\sin t - \dfrac{1}{2}Original equation
\displaystyle -\sin t\displaystyle =\displaystyle -\dfrac{1}{2}Subtract 2\sin t from both sides
\displaystyle \sin t\displaystyle =\displaystyle \dfrac{1}{2}Divide both sides by 4
\displaystyle t\displaystyle =\displaystyle \sin^{-1}\left(\dfrac{1}{2}\right)Take the inverse of \cos on both sides
\displaystyle =\displaystyle \dfrac{\pi}{6}Evaluate

To find the second quadrant solution:

\pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}

We can give the solution as t=\dfrac{\pi}{6},\,\dfrac{5\pi}{6}

Reflect and check

The solutions can be visualized by means of the following graphical representation.

The image shows curves that crosses in the x-axis. Ask your teacher for more information.

When there are several trigonometric terms involving different multiples of the variable and different trigonometric functions, it is usually necessary to use identities to make simplifications. For example, if there is a term in 2\theta as well as a term in \theta, we need to express both as functions of \theta. If there is a sine term and a cosine term, we would look for a way to express both with a single trigonometric function.

Example 7

Consider the equation \cos \theta = -0.798\,6.

a

Find the measure in degrees of the acute angle satisfying \cos \theta = 0.798\,6.

Round to the nearest degree.

Worked Solution
Create a strategy

We can use the inverse trigonometric function.

Apply the idea
\displaystyle \theta\displaystyle =\displaystyle \cos^{-1}(0.798\,6)Take the inverse of \cos on both sides
\displaystyle =\displaystyle 37\degreeEvaluate
b

Find the measure in degrees of the angles satisfying \cos \theta = - 0.798\,6 for 0\degree \leq \theta \leq 360\degree.

Round to the nearest degree.

Worked Solution
Create a strategy

Remember the value of \cos \theta is negative for an angle with measure \theta in the second and third quadrants.

Apply the idea

To find the solution for second quadrant, we need to subtract the value \theta from part (a) from 180\degree:

\displaystyle \theta\displaystyle =\displaystyle 180\degree - 37 \degreeSubtract 37 from 180
\displaystyle =\displaystyle 143\degreeEvaluate

To find the solution for third quadrant, we need to add the value \theta from part (a) from 180\degree:

\displaystyle \theta\displaystyle =\displaystyle 180\degree + 37\degreeAdd 37 to 180
\displaystyle =\displaystyle 217\degreeEvaluate

The solutions are \theta = 143\degree,\,217\degree.

Idea summary

We can determine the sign of the six basic trigonometric functions by relating the point \left( x, y \right) on the unit circle to the trigonometric functions \left ( \cos \theta , \sin \theta \right).

Outcomes

F.TF.B.7 (+)

Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context.

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