7.05 Trigonometric equations

Solve trigonometric equations

As we saw when looking at inverse trigonometric functions, we can use the inverse functions to solve one step trigonometric functions like $\sin\theta=\frac{1}{2}$sinθ=12​, by taking the applying the inverse ratio to both sides to give $\theta=\sin^{-1}\left(\frac{1}{2}\right)$θ=sin−1(12​).

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

 

Worked example

Question 1

Find $\theta$θ in degrees such that $2\tan^2\theta-1=49$2tan2θ−1=49, between $-90^\circ$−90° and $90^\circ$90°. Round to the nearest hundredth place.

Think: Since $\tan\theta$tanθ is not isolated, use standard algebraic techniques to isolate.  Remember, the inverse of a squared term is the square root of the term.

Do: We must make sure the calculator is set in degrees. We write

$2\tan^2\theta-1$2tan2θ−1	$=$=	$49$49
$2\tan^2\theta$2tan2θ	$=$=	$50$50
$\tan^2(\theta)$tan2(θ)	$=$=	$25$25
$\tan\theta$tanθ	$=$=	$\pm5$±5
$\theta$θ	$=$=	$\tan^{-1}(\pm5)$tan−1(±5)
$\theta$θ	$=$=	$\pm78.69^\circ$±78.69°

This is $\approx78.69^\circ$≈78.69° and $\approx-78.69^\circ$≈−78.69°

 

Solve trigonometric equations using reciprocal functions

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

$\operatorname{cosec}\theta$cosecθ	$=$=	$\frac{1}{\sin\theta}$1sinθ​
$\sec\theta$secθ	$=$=	$\frac{1}{\cos\theta}$1cosθ​
$\cot\theta$cotθ	$=$=	$\frac{\cos\theta}{\sin\theta}$cosθsinθ​	$=$=	$\frac{1}{\tan\theta}$1tanθ​

Worked examples

Question 2

Find the measure in degrees of the acute angle satisfying the equation $6\sec\theta+4=8\sec\theta$6secθ+4=8secθ

Think: We wish to solve for $\theta$θ. We can first aim to rewrite the equation with $\sec\theta$secθ as the subject, and then $\cos\theta$cosθ.

Do: 

$6\sec\theta+4$6secθ+4	$=$=	$8\sec\theta$8secθ	(Given)
$4$4	$=$=	$2\sec\theta$2secθ$\text{ }\text{ }\text{ }$	(Subtract $6\sec\theta$6secθ from both sides)
$2$2	$=$=	$\sec\theta$secθ	(Divide both sides by $2$2)
$2$2	$=$=	$\frac{1}{\cos\theta}$1cosθ​	(Substitute $\sec\theta=\frac{1}{\cos\theta}$secθ=1cosθ​)
$\frac{1}{2}$12​	$=$=	$\cos\theta$cosθ	(Reciprocate both sides)
$\theta$θ	$=$=	$\cos^{-1}\left(\frac{1}{2}\right)$cos−1(12​)	(Taking the inverse of $\cos$cos on both sides)

Evaluating this in the calculator, or referring to the exact value triangles, will show that $\theta=60^\circ$θ=60°.

 

Question 3

Find the measure in radians of the acute angle satisfying the equation $\sqrt{3}-2\cot\theta=\cot\theta$√3−2cotθ=cotθ.

Think: We wish to solve for $\theta$θ. We can first aim to rewrite the equation with $\cot\theta$cotθ as the subject, and then $\tan\theta$tanθ.

Do: 

$\sqrt{3}-2\cot\theta$√3−2cotθ	$=$=	$\cot\theta$cotθ	(Given)
$\sqrt{3}$√3	$=$=	$3\cot\theta$3cotθ	(Add $2\cot\theta$2cotθ from both sides)
$\frac{\sqrt{3}}{3}$√33​	$=$=	$\cot\theta$cotθ	(Divide both sides by $3$3)
$\frac{\sqrt{3}}{3}$√33​	$=$=	$\frac{1}{\tan\theta}$1tanθ​	(Substitute $\cot\theta=\frac{1}{\tan\theta}$cotθ=1tanθ​)
$\frac{3}{\sqrt{3}}$3√3​	$=$=	$\tan\theta$tanθ	(Reciprocate both sides)
$\theta$θ	$=$=	$\tan^{-1}\frac{3}{\sqrt{3}}$tan−13√3​	(Taking the inverse of $\tan$tan on both sides)
$\theta$θ	$=$=	$\frac{\pi}{6}$π6​	(Using exact values or your calculator)

 

Practice questions

Question 4

Question 5

 

Solve trigonometric equations using unit circle and exact values

Trigonometric equations express the situation in which we know the value of some trigonometric function, but we do not know what angle leads to that value. For example, we may know that for some angle $\theta$θ, we have $\sin\theta=\frac{\sqrt{3}}{2}$sinθ=√32​ or $\cos\beta=-\frac{1}{2}$cosβ=−12​. The problem is to determine what values of $\theta$θ and $\beta$β will make these true statements.

As we have seen previously, if we have a unit circle diagram, we can use the fact that for any point on the unit circle:

$x$x	$=$=	$\cos\theta$cosθ
$y$y	$=$=	$\sin\theta$sinθ
$\frac{y}{x}$yx​	$=$=	$\tan\theta$tanθ

 

Using the blue unit circle below, there are two instances where the $\sin$sin function is equal to $\frac{\sqrt{3}}{2}$√32​ ($60^\circ$60° and $120^\circ$120°)and two where $\cos$cos is equal to $-\frac{1}{2}$−12​ ($120^\circ$120° and $240^\circ$240°).

 

 

Worked example

Question 6

Using the unit circle diagram find all the solutions that lie between $0^\circ$0° and $360^\circ$360° of the equation $\sin2\theta=\frac{\sqrt{3}}{2}$sin2θ=√32​. 

Think: Recall $\frac{\sqrt{3}}{2}$√32​ represents $60^\circ$60° for $\sin\theta$sinθ , therefore we have $2\theta=60^\circ$2θ=60°. It must also be true that $2\theta=120^\circ$2θ=120° gives a second quadrant solution for $2\theta$2θ. We need to go around the unit circle again and find the additional solutions $2\theta=420^\circ$2θ=420° and $2\theta=480^\circ$2θ=480°.

Do:

Finally, by dividing by $2$2, we obtain the solutions for $\theta:$θ: $\theta=30^\circ$θ=30°, $\theta=60^\circ$θ=60°, $\theta=210^\circ$θ=210° and $\theta=240^\circ.$θ=240°.

 

Practice questions

Question 7

Question 8

Question 9

 

 

Solve further trigonometric equations across 4 quadrants for any angle 

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. 

Worked examples

Question 10

Given the function defined by $2\sin x=1$2sinx=1 for $x\in[0,360^\circ]$x∈[0,360°], find the values of $x$x .

Think: Notice the angle measure in the question is written in degrees. We must use check the calculator mode setting is on degrees and the we use the degrees indicated on the unit circle. The problem is to solve the equation $2\sin x=1$2sinx=1, giving the solutions that are between $0$0 and $360^\circ$360°. We must use algebraic techniques to isolate the trigonometric function. Then, use the inverse of the $\sin$sin function to find the solution in the between $0^\circ$0° and $90^\circ$90°. To find the remaining solutions, we will use $\sin\left(180^\circ-x\right)$sin(180°−x) 

Do:

$2\sin x$2sinx	$=$=	$1$1
$\sin x$sinx	$=$=	$\frac{1}{2}$12​
$x$x	$=$=	$\sin^{-1}\left(\frac{1}{2}\right)$sin−1(12​)
$x$x	$=$=	$30^\circ$30°

 

Or from previous work, we recall that $\sin30^\circ=\frac{1}{2}$sin30°=12​ , therefore $x=30^\circ$x=30°.

To find all solutions between $0^\circ$0°  and $360^\circ$360°, we must subtract the value of $x$x from $180^\circ$180°:

$\sin\left(180^\circ-30^\circ\right)=\sin150^\circ=\frac{1}{2}$sin(180°−30°)=sin150°=12​. So, these two are the only values that satisfy the equation.

We can give the solution as $x=30^\circ$x=30°, $150^\circ$150°.

Reflect: Another way to consider the solution is to think of $2$2 functions $y=2\sin x$y=2sinx and $y=1$y=1.  To solve $2\sin x=1$2sinx=1, is the same as graphically finding the intersection of these two separate lines.  

You can see both solutions mentioned above here.  

 

Question 11

Solve $4\cos x+1=3$4cosx+1=3 for $x\in[0,360^\circ]$x∈[0,360°].

Think: Notice the angle measure in the question is written in degrees. We must use check the calculator mode setting is on degrees and the we use the degrees indicated on the unit circle. Since the trigonometric function is not isolated, we must use algebraic techniques to isolate the function.  

Do:

$4\cos x+1$4cosx+1	$=$=	$3$3
$4\cos x$4cosx	$=$=	$2$2
$\cos x$cosx	$=$=	$\frac{1}{2}$12​
$x$x	$=$=	$\cos^{-1}\left(\frac{1}{2}\right)$cos−1(12​)

Using the unit circle or special triangles, we know that $x=60^\circ$x=60° and it is within the domain $[0,90^\circ]$[0,90°] . Again, from previous work, we know that $\cos x=\frac{1}{2}$cosx=12​ when $x=60^\circ$x=60°. The next solution must occur when $x=360-60$x=360−60. That is, when $x=300^\circ$x=300°.

Thus, the solutions are $x=60^\circ,300^\circ$x=60°,300°.

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

 

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

 

question 12

Find all the solutions of the equation $\sin t=2\sin t-\frac{1}{2}$sint=2sint−12​ in the interval $[0,2\pi]$[0,2π].

Think: Notice the angle measure in the question is written in degrees. We must use check the calculator mode setting is on radians and the we use the radians indicated on the unit circle. We must isolate the trigonometric function using algebraic techniques. Review the unit circle to determine with the solution will be an exact value or an approximation. 

Do:

$\sin t$sint	$=$=	$2\sin t-\frac{1}{2}$2sint−12​
$-\sin t$−sint	$=$=	$-\frac{1}{2}$−12​
$\sin t$sint	$=$=	$\frac{1}{2}$12​
$t$t	$=$=	$\sin^{-1}\left(\frac{1}{2}\right)$sin−1(12​)

 

Using the unit circle or special triangles, we find, in the first quadrant,  $t=\frac{\pi}{6}$t=π6​. To find the second quadrant solution, we subtract the first quadrant solution from $\pi$π. Thus, the second solution is $\frac{5\pi}{6}$5π6​.

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

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$2θ as well as a term in $\theta$θ, we need to express both as functions of $\theta$θ. If there is a sine term and a cosine term, we would look for a way to express both with a single trigonometric function.

Practice questions

Question 13

Question 14