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1.045 Trigonometric equations

Lesson

Trigonometric functions have many applications in periodic phenomena such as tides, pendulums, certain animal populations and analysing markets with seasonal changes. Just as with exponential functions before where we looked at some applications in further detail, let's look at how to solve some trigonometric equations.

 

Solving trigonometric equations

To find unknown angles in a right-angled triangles we can use the functions $\sin^{-1}$sin1, $\cos^{-1}$cos1 and $\tan^{-1}$tan1. When looking to find unknown angles in problems with trigonometric functions defined for angles beyond $90^\circ$90° or $\frac{\pi}{2}$π2, the difficulty 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=2sinx1. 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(xπ3)=1 for the interval $-2\pi\le x\le2\pi$2πx2π.

Think: 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(xπ3) 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(xπ3) (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(xπ3)=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

Calculator tips
  1. Ensure your calculator is in the correct mode of either radians or degrees, depending on the question.
  2. Using knowledge of transformations that have occurred and the interval given for solutions, select an appropriate viewing window before trying to find points of intersection.
  3. The calculator may not give the points of intersection in the exact form. Some problems may not have 'nice' solutions and sometimes we only need the answer correct to a certain number of decimal places. However, if we do require exact answers and you suspect the solutions are multiples of $\pi$π, you can find exact form by using the solve function of a CAS calculator, the numerical solve function in most graphics calculators or by dividing the $x$x-coordinate of the point of intersection by $\pi$π in either calculator to find what multiple of $\pi$π the solutions are appearing at.

 

Algebraic solutions

We can find unknown angles using our calculator with the functions $\sin^{-1}$sin1, $\cos^{-1}$cos1 and $\tan^{-1}$tan1, however, this only returns one solution. Which solution does it return and how do we find others?

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 $\cos^{-1}$cos1 or an angle in the fourth quadrant for $\sin^{-1}$sin1 and $\tan^{-1}$tan1.

Knowing the first solution, we can find others using the symmetry and period of the function.

 

Solving $\sin\left(bx-c\right)=a$sin(bxc)=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=sin1(a) and $\theta_2=\pi-\sin^{-1}\left(a\right)$θ2=πsin1(a).

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

$bx-c$bxc $=$= $\theta_1+2\pi k$θ1+2πk or $bx-c$bxc $=$= $\theta_2+2\pi k$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(bxc)=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=cos1(a) and $\theta_2=-\cos^{-1}\left(a\right)$θ2=cos1(a).

 

Solving $\tan\left(bx-c\right)=a$tan(bxc)=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=tan1(a).

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

$bx-c$bxc $=$= $\theta_1+\pi k$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 examples

Example 2

Find solutions to $\sin x=0.8$sinx=0.8 for the domain $0\le x\le4\pi$0x4π. Round answers to three decimal places.

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

Do: Find first two solutions

$\sin x$sinx $=$= $0.8$0.8        
$x$x $=$= $\sin^{-1}\left(0.8\right)$sin1(0.8) or $x$x $=$= $\pi-\sin^{-1}\left(0.8\right)$πsin1(0.8)
  $=$= $0.927\dots$0.927     $=$= $2.214\dots$2.214

Hence, all possible solutions are of the form $x=0.927\dots+2\pi k$x=0.927+2πk or $x=2.214\dots+2\pi k$x=2.214+2πk, for $k$k any integer. Since we want solutions in the interval $0\le x\le4\pi$0x4π, 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$0x4π are: $0.927$0.927, $2.214$2.214, $4.069$4.069 and $8.497$8.497.

 

Example 3

Find solutions to $2\cos x+1=0$2cosx+1=0 on the domain $-2\pi\le x\le2\pi$2πx2π.

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 the figure below, 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$0x2π, to find ones for the domain $-2\pi\le x\le2\pi$2πx2π we can subtract $2\pi$2π from each of these solutions. Hence, the full set of solutions we are seeking are:

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

Example 4

Find solutions to $2\cos\left(3x-\frac{\pi}{3}\right)-\sqrt{3}=0$2cos(3xπ3)3=0 for the interval $0\le x\le2\pi$0x2π.

Think: First rearrange:

$\cos\left(3x-\frac{\pi}{3}\right)$cos(3xπ3) $=$= $\frac{\sqrt{3}}{2}$32

Next, find initial solutions.

Do: Since $\frac{\sqrt{3}}{2}$32 is a familiar exact value, we can find our initial solutions using the blue diagram below. 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}$3xπ3 $=$= $\frac{\pi}{6}+2\pi k$π6+2πk or $3x-\frac{\pi}{3}$3xπ3 $=$= $\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 or subtract multiples of $\frac{2\pi}{3}$2π3 and only consider those within the interval $0\le x\le2\pi$0x2π.

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.

 

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 for equations involving $\sin\theta$sinθ or $\cos\theta$cosθ and the values the values: $0$0, $\pm1$±1, $\pm\sqrt{3}$±3, $\pm\frac{1}{\sqrt{3}}$±13 for equations involving $\tan\theta$tanθ. 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)

 

Tips

Some equations may make use of identities or factorisation before solving. Examples:

  • For an equation such as $\sin(5x)=\cos(5x)$sin(5x)=cos(5x), we can divide both sides by $\cos(5x)$cos(5x) and then make use of the identity $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ to reduce the equation to $\tan(5x)=1$tan(5x)=1 before solving.
  • For an equation such as $2\cos^2x+3\sin^2x=1$2cos2x+3sin2x=1 we can make use of the identity $\sin^2\theta+\cos^2\theta=1$sin2θ+cos2θ=1 to reduce the equation to $2+\sin^2\theta=1$2+sin2θ=1 before rearranging and solving.
  • For an equation such as $2\sin^2x-\sin x-1=0$2sin2xsinx1=0, recognise the equation is a quadratic in terms of $\sin x$sinx and factorise to obtain $(2\sin x+1)(\sin x-1)=0$(2sinx+1)(sinx1)=0, then equate each factor to zero and solve.

 

Practice questions

question 1

Consider the function $y=3\sin x$y=3sinx.

  1. Graph this function.

    Loading Graph...

  2. Add the line $y=3$y=3 to your graph.

    Loading Graph...
  3. Hence, state all solutions to the equation $3\sin x=3$3sinx=3 over the domain $\left[-2\pi,2\pi\right]$[2π,2π]. Give your answers as exact values separated by commas.

question 2

Consider the functions $y=-\cos\left(x-\frac{\pi}{4}\right)-2$y=cos(xπ4)2 and $y=-3$y=3.

  1. Draw the functions $y=-\cos\left(x-\frac{\pi}{4}\right)-2$y=cos(xπ4)2 and $y=-3$y=3.

    Loading Graph...

  2. Hence, state all solutions to the equation $-\cos\left(x-\frac{\pi}{4}\right)-2=-3$cos(xπ4)2=3 over the domain $\left(-2\pi,2\pi\right)$(2π,2π). Give your answers as exact values separated by commas.

question 3

Consider the function $y=\tan\left(x-\frac{\pi}{4}\right)$y=tan(xπ4).

  1. Graph this function.

    Loading Graph...

  2. Add the line $y=1$y=1 to your graph.

    Loading Graph...
  3. Hence, state all solutions to the equation $\tan\left(x-\frac{\pi}{4}\right)=1$tan(xπ4)=1 over the domain $\left[-2\pi,2\pi\right)$[2π,2π). Give your answers as exact values separated by commas.

Question 4

Find all values of $\theta$θ in the interval $[$[$0,2\pi$0,2π$)$) that satisfy $\sin\theta=\frac{1}{2}$sinθ=12.

Write all values on the same line separated by a comma.

Question 5

Solve $6\cos x-3\sqrt{2}=0$6cosx32=0 over the interval $\left[0,2\pi\right)$[0,2π).

Question 6

Solve $\sqrt{3}\tan\left(\frac{x}{2}\right)=-3$3tan(x2)=3 for $0\le x<2\pi$0x<2π.

Question 7

Solve for $x$x over the interval $[$[$0$0, $2\pi$2π$)$).

$\cos^2\left(\frac{x}{2}\right)-1=0$cos2(x2)1=0

  1. Enter each line of work as an equation.

Outcomes

ACMMM103

use trigonometric functions and their derivatives to solve practical problems

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