As we saw when looking at inverse trigonometric functions, we can use the inverse functions to solve one step trigonometric functions like \text{sin }\theta =\dfrac{1}{2}, by taking the applying the inverse ratio to both sides 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.

Examples

Example 1

Find \theta in degrees such that 2 \text{ tan}^{2 } \theta - 1 = 49, between -90\degree and 90\degree. Round to the nearest hundredth place.

Worked Solution

Create a strategy

Since \text{tan } \theta is not isolated, we will use standard algebraic techniques to isolate.

Apply the idea

\displaystyle 2 \text{ tan}^{2}\theta	\displaystyle =	\displaystyle 50	Add 1 to both sides
\displaystyle \text{ tan}^{2}(\theta)	\displaystyle =	\displaystyle 25	Divide both sides by 2
\displaystyle \text{ tan }\theta	\displaystyle =	\displaystyle \pm 5	Take the square root on both sides
\displaystyle \theta	\displaystyle =	\displaystyle \text{tan}^{-1}(\pm 5)	Take the inverse tangent
\displaystyle \theta	\displaystyle =	\displaystyle \pm 78.69\degree	Simplify

So we have \theta \approx 78.69\degree and \theta \approx -78.69\degree.

Idea summary

A trigonometric equation is an equation that involves one or more of the six trigonometric functions: sine \text{(sin)}, cosine \text{(cos)}, tangent \text{(tan)}, cosecant \text{(csc)}, secant \text{(sec)}, and cotangent \text{(cot)}.

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 2

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\degree	Evaluate

Example 3

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 4	Add 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\degree	Evaluate

Idea summary

Remember the relationship between the reciprocal trigonometric functions and 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}

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.

Recall 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 4

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

Trigonometric equations and inequalities

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 5

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 1	Original 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\degree	Evaluate

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 6

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

Worked Solution

Create a strategy

Start by isolating the trigonometric function on one side of the inequality. Then use a graph of the cosine function to identify which domain regions satisfy the inequality.

Apply the idea

\displaystyle 4\cos x + 1	\displaystyle >	\displaystyle 3	Original equation
\displaystyle 4\cos x	\displaystyle >	\displaystyle 2	Subtract 1 from both sides
\displaystyle \cos x	\displaystyle >	\displaystyle 0.5	Divide both sides by 4

Now let's look at a graph of y = \cos\left(x\right) and compare it to the value y = 0.5

180

270

360

-1

-0.5

0.5

We can find the points of intersection by taking the inverse cosine. This gives us the first quadrant solution of \cos^{-1}\left(0.5\right) = 60\degree, and the corresponding fourth quadrant solution is 360\degree - 60\degree = 300\degree.

Using the graph, we can see that \cos\left(x\right) > 0.5 is satisfied when the cosine curve lies above the line y = 0.5. This happens for x \in \left[0\degree, 60\degree\right) and x \in \left(300\degree, 360\degree\right].

Example 7

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

Worked Solution

Create a strategy

Start by isolating the trigonometric function on one side of the inequality, then solve the resulting equation for the critical points, and use the unit circle or a sketch of the sine function to determine the intervals on which the original inequality holds true.

Apply the idea

\displaystyle \sin t	\displaystyle \leq	\displaystyle 2\sin t - \dfrac{1}{2}	Original equation
\displaystyle 0	\displaystyle \leq	\displaystyle \sin t-\dfrac{1}{2}	Subtract \sin t from both sides
\displaystyle \dfrac{1}{2}	\displaystyle \leq	\displaystyle \sin t	Add \dfrac{1}{2} to both sides

We can now solve the equation \sin t = \dfrac{1}{2} for the critical points.

Taking the inverse sine of both sides, we get t = \sin^{-1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{6} as the first quadrant critical point. Since we want to consider all solutions in the interval \left[0,\, 2\pi\right], and sine is positive in the first two quadrants, we also want the second quadrant critical point: \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}.

We can now sketch a graph of y = \sin\left(t\right) and compare it to the value y = \dfrac{1}{2} to find the solution sets:

\frac{1}{2}\pi

1\pi

\frac{3}{2}\pi

2\pi

-1

-\frac{1}{2}

\frac{1}{2}

We see that the curve lies above the line in between the critical points of \dfrac{\pi}{6} and \dfrac{5\pi}{6}. So our solution set is t \in \left[\dfrac{\pi}{6}, \dfrac{5\pi}{6}\right].

Example 8

Consider the value 0.798\,6.

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\degree	Evaluate

Find the measure in degrees of the angles satisfying \cos \theta \lt - 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

We know the first quadrant solution of \cos \theta = 0.798\,6 from part (a), and we can use this to find the corresponding second and third quadrant critical points for \cos\left(\theta\right) = -0.7986, to the nearest degree:

\displaystyle \theta	\displaystyle =	\displaystyle 180\degree \pm 37\degree
	\displaystyle =	\displaystyle 143\degree, 217\degree

Since cosine is negative in the second and third quadrants, we know that the solutions to \cos\left(\theta\right) < -0.798\,6 must lie between these two critical points.

Therefore, the angles satisfying the condition are \theta \in \left(143\degree, 217\degree\right).

Example 9

Jack decides to take a spin on the carousel at the local fair. His distance from the center of the carousel (in meters) after t minutes is given by y\left(t\right) = -10 \cos \left(\dfrac{\pi t}{2} \right) + 12. Each ride includes four full rotations of the carousel.

How many minutes does one ride last?

Worked Solution

Create a strategy

Use the formula T = \dfrac{2\pi}{B}, where B is the coefficient of t in the sine function, and then multiply it by the number of complete rotations.

Apply the idea

\displaystyle T	\displaystyle =	\displaystyle \dfrac{2\pi}{\left(\dfrac{\pi}{2}\right)}	Substitute B = \dfrac{\pi}{2}
	\displaystyle =	\displaystyle 4	Evaluate

So one full rotation takes 4 minutes. Each ride includes 4 full rotations, and therefore one ride lasts for 4 \times 4 = 16 minutes.

Find all times t at which Jack is exactly 10 meters away from the center of the carousel.

Worked Solution

Create a strategy

We want to solve for the values of t when y\left(t\right) = 10. That is, we want to solve the equation -10 \cos \left(\dfrac{\pi t}{2} \right) + 12 = 10

Apply the idea

\displaystyle -10 \cos \left(\dfrac{\pi t}{2} \right) + 12	\displaystyle =	\displaystyle 10	Set y(t)=10
\displaystyle -10\cos \left(\dfrac{\pi t}{2}\right)	\displaystyle =	\displaystyle -2	Subtract 12 from both sides
\displaystyle \cos \left(\dfrac{\pi t}{2}\right)	\displaystyle =	\displaystyle 0.2	Divide both sides by -10

Using the inverse cosine function, we get a first quadrant solution of

\displaystyle \dfrac{\pi t}{2}	\displaystyle =	\displaystyle \cos^{-1}\left(0.2\right)
\displaystyle t	\displaystyle =	\displaystyle \dfrac{2\cos^{-1}\left(0.2\right)}{\pi}
	\displaystyle \approx	\displaystyle 0.872 \text{ min}

Since the cosine function is positive in the first and fourth quadrants, we also get a corresponding fourth quadrant solution of

\displaystyle \dfrac{\pi t}{2}	\displaystyle =	\displaystyle 2\pi - \cos^{-1}\left(0.2\right)
\displaystyle t	\displaystyle =	\displaystyle \dfrac{2\left(2\pi - \cos^{-1}\left(0.2\right)\right)}{\pi}
	\displaystyle \approx	\displaystyle 3.128 \text{ min}

From part (a), we know that each rotation lasts for 4 minutes, and the total length of the ride is 16 minutes.

So the solutions will all be in the domain t \in \left[0, 16\right], and we can find the rest of the solutions by adding multiples of 4 minutes to our two current solutions.

Therefore, the solutions are:t = \dfrac{2\cos^{-1}\left(0.2\right)}{\pi} + 4n,\, \dfrac{2\left(2\pi - \cos^{-1}\left(0.2\right)\right)}{\pi} + 4n,\, \text{ for }n \in \left\{0, 1, 2, 3\right\}or, rounding to three decimal places:t \approx 0.872,\, 3.128,\, 4.872,\, 7.128,\, 8.872,\, 11.128,\, 12.872,\, 15.128 \text{ min}

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).

3.10 Trigonometric equations and inequalities

Introduction

Ideas

Solve trigonometric equations

Examples

Example 1

Solve using reciprocal functions

Examples

Example 2

Example 3

Solve using the unit circle

Examples

Example 4

Trigonometric equations and inequalities

Examples

Example 5

Example 6

Example 7

Example 8

Example 9

Outcomes

3.10.A

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