5.10 Trigonometric equations

In Chapter 4 we solved questions to find unknown angles in a right-angled triangles using the functions $\sin^{-1}$sin−1, $\cos^{-1}$cos−1 and $\tan^{-1}$tan−1. We now want to look at finding unknown angles in problems with our trigonometric function now defined for angles beyond $90^\circ$90° or $\frac{\pi}{2}$π2​. The difficulty here is there can be infinitely many solutions.

 

We will focus on cases where the domain in restricted, so there will be a finite number of solutions. But how do we find them? Let's look at two main methods: finding solutions graphically and finding solutions algebraically.

Graphical solutions

To solve an equation graphically, such as $2\sin x+1=0$2sinx+1=0 where the right-hand side is zero, we are essentially finding the $x$x-intercepts of the graph of $y=2\sin x-1$y=2sinx−1. If the right-hand side was not equal to zero, such as $2\sin x=-1$2sinx=−1, we can move all terms to the left hand side and again find the $x$x-intercepts of the graph. Alternatively, we can graph both sides of the equation as separate functions, for our example $y=2\sin x$y=2sinx and $y=-1$y=−1, and then find the $x$x-coordinates of points of intersection of the two curves. We can see these methods are equivalent, in fact the first method is simply finding the intersection of the curve with the line $y=0$y=0.

In general, solving an equation can be thought of as finding the $x$x-values of the points of intersection of two curves representing the left- and right-hand side of the equation.

Worked example

Example 1

Find the values of $x$x that solve the equation $\sin\left(x-\frac{\pi}{3}\right)=1$sin(x−π3​)=1 for the interval $-2\pi\le x\le2\pi$−2π≤x≤2π.

Graphically speaking, this is the same as finding the $x$x-coordinates that correspond to the points of intersection of the curves $y=\sin\left(x-\frac{\pi}{3}\right)$y=sin(x−π3​) and $y=1$y=1. Generally, we will use technology to solve such equations graphically. Graphing both functions using technology and taking care to show the correct domain we have:

$y=\sin\left(x-\frac{\pi}{3}\right)$y=sin(x−π3​) (green) and $y=1$y=1 (blue).

 

We can see in the region given by $\left(-2\pi,2\pi\right)$(−2π,2π) that there are two points where the two functions meet.

Points indicating where the two functions meet.

 

Since we are fortunate enough to have gridlines that coincide with the intersections, the $x$x-values for these points of intersection can be easily deduced. Each grid line is separated by $\frac{\pi}{6}$π6​, which means that the solution to the equation $\sin\left(x-\frac{\pi}{3}\right)=1$sin(x−π3​)=1 in the region $\left(-2\pi,2\pi\right)$(−2π,2π) is given by:

$x=-\frac{7\pi}{6},\frac{5\pi}{6}$x=−7π6​,5π6​

 

Practice questions

question 1

question 2

question 3

Algebraic solutions

We can find unknown angles using our calculator with the functions $\sin^{-1}$sin−1, $\cos^{-1}$cos−1 and $\tan^{-1}$tan−1, however, this only returns one solution. Which solution does it return and how do we find others? The diagrams below show the region the calculator will look for a solution for each function:

$y=\sin x$y=sinx
$y=\cos x$y=cosx
$y=\tan x$y=tanx

For a positive ratio these functions will always return an angle in the first quadrant and for a negative ratio it will return an angle in the second quadrant for $\cos^{-1}$cos−1 or an angle in the fourth quadrant for $\sin^{-1}$sin−1 and $\tan^{-1}$tan−1.

Knowing the first solution, we can find others using the symmetry of the function. Let's look at how to find solutions for equations of the form $\sin x=a$sinx=a, $\cos x=a$cosx=a and $\tan x=a$tanx=a

Solutions for $\sin x=a$sinx=a and $a$a is positive	Solutions for $\sin x=a$sinx=a and $a$a negative

Solving for $\sin x=a$sinx=a there will be two solutions for each period of the graph for any value of $a$a between the max and min. We can find our initial solution $x=\theta_1$x=θ1​ by using using our calculator to evaluate: $\theta_1=\sin^{-1}\left(a\right)$θ1​=sin−1(a).

Then the second solution $x=\theta_2$x=θ2​ can be found from the symmetry above: $\theta_2=\pi-\theta_1$θ2​=π−θ1​.

To find any more solutions we can add any integer multiple of the period. In this case the period is $2\pi$2π.

Solutions for $\cos x=a$cosx=a

Solving for $\cos x=a$cosx=a there will be two solutions for each period of the graph for any value of $a$a between the max and min. We can find our initial solution $x=\theta_1$x=θ1​ by using using our calculator to evaluate: $\theta_1=\cos^{-1}\left(a\right)$θ1​=cos−1(a).

Since, cosine is an odd function, the second solution $x=\theta_2$x=θ2​ can be found from the symmetry: $\theta_2=-\theta_1$θ2​=−θ1​.

To find any more solutions we can add any integer multiple of the period. In this case the period is $2\pi$2π.

Solutions for $\tan x=a$tanx=a

Solving for $\tan x=a$tanx=a there will be a single solution for each period of the graph for any value of $a$a. We can find our initial solution $x=\theta_1$x=θ1​ by using using our calculator to evaluate: $\theta_1=\tan^{-1}\left(a\right)$θ1​=tan−1(a).

To find any more solutions we can add any integer multiple of the period. In this case the period is $\pi$π.

Reminder: some angles give exact values when evaluated using a trigonometric function. You should recognise the values: $0$0, $\pm1$±1, $\pm\frac{1}{2}$±12​, $\pm\frac{\sqrt{2}}{2}$±√22​ and $\pm\frac{\sqrt{3}}{2}$±√32​ and while you can use the calculator to find associated angles you could also use a table or diagrams, such as those below, to find solutions. This method may be more efficient once practiced and may be required for assessments without a calculator.

Multiples of $45^\circ$45° $\left(\frac{\pi}{4}\right)$(π4​)	Multiples of $30^\circ$30° $\left(\frac{\pi}{6}\right)$(π6​)

 

Worked example

Example 2

Find solutions to $\sin x=0.8$sinx=0.8 for the domain $0\le x\le4\pi$0≤x≤4π. Give your answers correct to three decimal places.

Think: Is $0.8$0.8 a familiar value from an exact value table or diagram? No - so we proceed with using calculator.

Do: Find first two solutions

$\sin x$sinx	$=$=	$0.8$0.8
$x$x	$=$=	$\sin^{-1}\left(0.8\right)$sin−1(0.8)	or	$x$x	$=$=	$\pi-\sin^{-1}\left(0.8\right)$π−sin−1(0.8)
	$=$=	$0.927$0.927 ($3$3 decimal places)			$=$=	$2.214$2.214 ($3$3 decimal places)

Hence, all possible solutions are of the form $x=0.927+2\pi k$x=0.927+2πk or $x=2.214+2\pi k$x=2.214+2πk, for $k$k any integer. Since we want solutions in the interval $0\le x\le4\pi$0≤x≤4π, we need our initial solutions plus two more with one period ($2\pi$2π) added to each.

So solutions for $\sin x=0.8$sinx=0.8 on the domain $0\le x\le4\pi$0≤x≤4π are: $0.927$0.927, $2.214$2.214, $7.210$7.210 and $8.497$8.497, to three decimal places.

Example 3

Find solutions to $2\cos x+1=0$2cosx+1=0 on the domain $-2\pi\le x\le2\pi$−2π≤x≤2π.

Think: First rearrange the equation to the form $\cos x=a$cosx=a.

$2\cos x+1$2cosx+1	$=$=	$0$0
$2\cos x$2cosx	$=$=	$-1$−1
$\cos x$cosx	$=$=	$\frac{-1}{2}$−12​

Since $\frac{-1}{2}$−12​ is one of our values we should recognise we can proceed with either the calculator or diagram.

Do: Using the blue exact value diagram from above we look for the angles which show an $x$x-coordinate of $\frac{-1}{2}$−12​. We see that the angles $\frac{2\pi}{3}$2π3​ and $\frac{4\pi}{3}$4π3​ are solutions. These are the solutions for $0\le x\le2\pi$0≤x≤2π, to find ones for the domain $-2\pi\le x\le2\pi$−2π≤x≤2π we can subtract $2\pi$2π from each of these solutions. Hence, the full set of solutions we are seeking is:

$\frac{2\pi}{3}$2π3​, $\frac{4\pi}{3}$4π3​, $\frac{-2\pi}{3}$−2π3​ and $\frac{-4\pi}{3}$−4π3​.

Practice questions

Question 4

Question 5

 

Further trigonometric equations

We can solve equations with more steps by rearranging the equation to a familiar form and then using reverse operations to isolate the unknown. Let's look at general steps we can take to solve equations of the form $\sin\left(bx-c\right)=a$sin(bx−c)=a, $\cos\left(bx-c\right)=a$cos(bx−c)=a and $\tan\left(bx-c\right)=a$tan(bx−c)=a.

Solving $\sin\left(bx-c\right)=a$sin(bx−c)=a:

First find the initial solutions $\theta_1$θ1​ and $\theta_2$θ2​ for the equation $\sin\theta=a$sinθ=a. This can be found using a diagram for exact values or your calculator. $\theta_1=\sin^{-1}\left(a\right)$θ1​=sin−1(a) and $\theta_2=\pi-\sin^{-1}\left(a\right)$θ2​=π−sin−1(a).

Hence, all solutions to the equation can be found by rearranging:

$bx-c$bx−c

$=$=

$\theta_1+2\pi k$θ1​+2πk

or

$bx-c$bx−c

$=$=

$\theta_2+2\pi k$θ2​+2πk

, where $k$k is an integer.

By adding $c$c to both sides and dividing by $b$b, we obtain:

$x$x

$=$=

$\frac{\theta_1+c}{b}+\frac{2\pi k}{b}$θ1​+cb​+2πkb​

or

$x$x

$=$=

$\frac{\theta_2+c}{b}+\frac{2\pi k}{b}$θ2​+cb​+2πkb​

, where $k$k is an integer.

Final step is to find solutions in the interval given. Let $k$k be appropriate integer values to achieve the solutions in the given range.

 

Solving $\cos\left(bx-c\right)=a$cos(bx−c)=a:

The steps for solving an equation involving cosine are the same as above except for finding the initial solutions. The solutions $\theta_1$θ1​ and $\theta_2$θ2​ for the equation $\cos\theta=a$cosθ=a can be found using a diagram for exact values or your calculator. $\theta_1=\cos^{-1}\left(a\right)$θ1​=cos−1(a) and $\theta_2=-\cos^{-1}\left(a\right)$θ2​=−cos−1(a).

 

Solving $\tan\left(bx-c\right)=a$tan(bx−c)=a:

First find the initial solution, $\theta_1$θ1​, for tan there is a single solution for each period. The solution $\theta_1$θ1​ for the equation $\tan\theta=a$tanθ=a can be found using a diagram for exact values or your calculator. $\theta_1=\tan^{-1}\left(a\right)$θ1​=tan−1(a).

Hence, all solutions to the equation can be found by rearranging:

$bx-c$bx−c

$=$=

$\theta_1+\pi k$θ1​+πk

, where $k$k is an integer.

Remember $\tan x$tanx has a period of $\pi$π.

By adding $c$c to both sides and dividing by $b$b, we obtain:

$x$x

$=$=

$\frac{\theta_1+c}{b}+\frac{\pi k}{b}$θ1​+cb​+πkb​

, where $k$k is an integer.

Final step is to find solutions in the interval given. Let $k$k be appropriate integer values to achieve the solutions in the given range.

Worked example

Example 4

Find solutions to $2\cos\left(3x-\frac{\pi}{3}\right)-\sqrt{3}=0$2cos(3x−π3​)−√3=0 for the interval $0\le x\le2\pi$0≤x≤2π.

First rearrange:

$\cos\left(3x-\frac{\pi}{3}\right)$cos(3x−π3​)

$=$=

$\frac{\sqrt{3}}{2}$√32​

Next, find initial solutions. Since $\frac{\sqrt{3}}{2}$√32​ is a familiar exact value, we can find our initial solutions using the blue diagram from earlier in this lesson. Finding values which give an $x$x-coordinate of $\frac{\sqrt{3}}{2}$√32​, we find $\theta_1=\frac{\pi}{6}$θ1​=π6​ and $\theta_2=\frac{11\pi}{6}$θ2​=11π6​.

Hence, all possible solutions are:

$3x-\frac{\pi}{3}$3x−π3​	$=$=	$\frac{\pi}{6}+2\pi k$π6​+2πk	or	$3x-\frac{\pi}{3}$3x−π3​	$=$=	$\frac{11\pi}{6}+2\pi k$11π6​+2πk	, where $k$k is an integer.
$3x$3x	$=$=	$\frac{\pi}{2}+2\pi k$π2​+2πk		$3x$3x	$=$=	$\frac{13\pi}{6}+2\pi k$13π6​+2πk	, where $k$k is an integer.
$x$x	$=$=	$\frac{\pi}{6}+\frac{2\pi k}{3}$π6​+2πk3​		$x$x	$=$=	$\frac{13\pi}{18}+\frac{2\pi k}{3}$13π18​+2πk3​	, where $k$k is an integer.

Since we want solutions up to $2\pi$2π we can add the $\frac{2\pi}{3}$2π3​ zero, once or twice to the first solution and for the second solution we can add it zero, once or subtract it once to stay within the interval.

So our final set of solutions is: $\frac{\pi}{18}$π18​, $\frac{\pi}{6}$π6​, $\frac{13\pi}{18}$13π18​, $\frac{5\pi}{6}$5π6​, $\frac{25\pi}{18}$25π18​ and $\frac{3\pi}{2}$3π2​.

Practice questions

Question 6

Question 7

Question 8