6.03 Solutions to trigonometric equations

There are two methods available to us for solving trigonometric equations. The algebraic method requires that the equations are in a form, or can be manipulated into a form, that we can solve. The graphical method of solving equations proves very useful in determining solutions to complex trigonometric equations that may be difficult to solve algebraically. When used in conjunction with technology, the graphical method allows us to get very good estimates of the solutions.

Degrees and radians

When we first learned about trigonometry, we thought about angles in terms of degrees. We also used degrees to solve simpler trigonometric equations. Angles can be measured in degrees, gradians or radians. For example, a $90$90 degrees angle is equivalent to $100$100 gradians and $\frac{\pi}{2}$π2​ radians. Degrees and gradians are somewhat arbitrary units that make mathematics easier in some contexts.

Radians, on the other hand, relate an angle measure to a length and hence ensure that, from the perspective of the graphical solution of trigonometric equations, both the scales on the $x$x and $y$y-axis are congruent. This is vital when solving equations involving both trigonometric and other functions. For example, for the equation $\sin(x-\pi)=e^{-x}$sin(x−π)=e−x, plotting the exponential function with $x$x in degrees would not make sense.

Algebraic solutions to trigonometric equations

In this section we will look at strategies for algebraically solving trigonometric equations involving functions in the form $kf(a(x+b))+c$kf(a(x+b))+c. We need to take care to ensure appropriate adjustments are made to the specified domain during the process of our solution.

Worked example

Example 1

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.

Think: We will first rearrange the equation into a form that we can solve and help us determine the adjustments that need to be made to the domain.

Do: Rearranging the equation:

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

$=$=

$\frac{\sqrt{3}}{2}$√32​

 

The argument of the cos function is $3x-\frac{\pi}{3}$3x−π3​, hence we need to multiple the domain by $3$3 and subtract $\frac{\pi}{3}$π3​ to give the new domain for the rearranged equation above.

Multiplying the domain by $3$3 gives:

$0\le3x\le6\pi$0≤3x≤6π

Subtracting $\frac{\pi}{3}$π3​ from all parts of the inequality gives:

$-\frac{\pi}{3}\le3x-\frac{\pi}{3}\le\frac{17\pi}{3}$−π3​≤3x−π3​≤17π3​

Next, we solve $\cos(3x-\frac{\pi}{3})=\frac{\sqrt{3}}{2}$cos(3x−π3​)=√32​ using our normal methods but over the new domain. Hence we think, $\cos$cos of what angle gives $\frac{\sqrt{3}}{2}$√32​? As $\frac{\sqrt{3}}{2}$√32​ is an exact value and is positive we can use the unit circle to determine the solutions.

Determining an initial solution to our equation in the first quadrant, we can see that $\cos$cos of the angle $\frac{\pi}{6}$π6​ gives $\frac{\sqrt{3}}{2}$√32​ in the first quadrant and we also know that $\cos$cos is positive in the fourth quadrant.

Starting at the minimum value of our domain $-\frac{\pi}{3}$−π3​, which is equivalent to an angle of $\frac{5\pi}{3}$5π3​ shown on our diagram above, we can see that our first solution will be $-\frac{\pi}{6}$−π6​ (equivalent to $\frac{11\pi}{6}$11π6​ on our diagram above). Note that this is given as a negative angle because the first part of our domain is negative. Moving anti-clockwise around the circle from here up to the maximum point of our domain of $\frac{17\pi}{3}$17π3​, and only including solutions that give positive $\frac{\sqrt{3}}{2}$√32​, we get:

$3x-\frac{\pi}{3}=-\frac{\pi}{6},\frac{\pi}{6},\frac{11\pi}{6},\frac{13\pi}{6},\frac{23\pi}{6},\frac{25\pi}{6}$3x−π3​=−π6​,π6​,11π6​,13π6​,23π6​,25π6​

Now we solve this equation by first adding $\frac{\pi}{3}$π3​ to each solution giving:

$3x=\frac{\pi}{6},\frac{\pi}{2},\frac{13\pi}{6},\frac{15\pi}{6},\frac{25\pi}{6},\frac{27\pi}{6}$3x=π6​,π2​,13π6​,15π6​,25π6​,27π6​

Dividing both sides of the equation by $3$3 gives the solutions to our equations as follows:

$x=\frac{\pi}{18},\frac{\pi}{6},\frac{13\pi}{18},\frac{5\pi}{6},\frac{25\pi}{18},\frac{3\pi}{2}$x=π18​,π6​,13π18​,5π6​,25π18​,3π2​

It is good practice to do a check to make sure all our $x$x values lie within our original domain of $0\le x\le2\pi$0≤x≤2π which they do in this case.

Practice questions

Question 1

Question 2

Graphical solutions to trigonometric equations

The graphical solution to trigonometric equations requires that both sides of the equation be considered as separate functions. These functions are then plotted and the intersection points between the two functions are the solutions to the equation. The simpler equations can be solved in either degrees or radians if one of the functions is a constant value or both sides of the equation involve a trigonometric function. However, remember, if there is a trigonometric function and another type of function, the trigonometric function will need to be considered in radians to ensure equality of axis scales.

Worked examples

Example 2

Find all the solutions to the equation $\sin\left(x-60^\circ\right)=1$sin(x−60°)=1 over the domain of $(-360,360).$(−360,360).

Think: Graphically speaking, this is the same as finding the $x$x-values that correspond to the points of intersection of the curves $y=\sin\left(x-60^\circ\right)$y=sin(x−60°) and $y=1$y=1. As we are working with a constant value for one of the functions we can work in degrees for this problem.

Do: Plotting both graphs:

$y=\sin\left(x-60^\circ\right)$y=sin(x−60°) (green) and $y=1$y=1 (blue).

 

We can see in the region given by $\left(-360^\circ,360^\circ\right)$(−360°,360°) 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, the $x$x-values for these points of intersection can be easily deduced. Each grid line is separated by $30^\circ$30°, which means that the solution to the equation $\sin\left(x-60^\circ\right)=1$sin(x−60°)=1 in the region $\left(-360^\circ,360^\circ\right)$(−360°,360°) is given by:

$x=-210^\circ,150^\circ$x=−210°,150°

Example 3

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.

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(x−π3​) and $y=1$y=1. Generally, we will use technology to solve such equations graphically.

Do: 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​

Example 4

Given that $\tan x=\sin3x+5$tanx=sin3x+5, find all the values of $x$x that satisfy this equation between $0^\circ$0° and $360^\circ$360°.

Think: As the solution interval is provided in degrees, it is important that we provide our answers in the same units. This equation is difficult to solve algebraically, and hence is a great candidate for a graphical solution. Note that we can solve this question graphically in degrees because both sides of the equation can be plotted using this unit.

Do: In the image below we have drawn the graphs of two functions, $f\left(x\right)=\tan x$f(x)=tanx and $g\left(x\right)=\sin3x+5$g(x)=sin3x+5. The region $0^\circ\le x\le360^\circ$0°≤x≤360° has been shaded, since we are only concerned about the solutions in this interval.

$f\left(x\right)=\tan x$f(x)=tanx (green) and $g\left(x\right)=\sin3x+5$g(x)=sin3x+5 (blue).

 

We can see that there are two points of intersection where $f\left(x\right)=g\left(x\right)$f(x)=g(x), and the $x$x-coordinates of these two points correspond to the solutions to the equation $\tan x=\sin3x+5$tanx=sin3x+5.

Using appropriate technology, we determine that the points of intersection are $\left(76.707^\circ,4.233\right)$(76.707°,4.233) and $\left(260.340^\circ,5.875\right)$(260.340°,5.875). So the solutions to the equation $\tan x=\sin3x+5$tanx=sin3x+5 are $x=77^\circ$x=77° and $x=260^\circ$x=260°, to the nearest degree.

Example 5

Using technology, determine the solutions to $e^{x-4}=\sin(2x)$ex−4=sin(2x). over the interval $[0,2\pi]$[0,2π].

Think: On appropriate technology, we could plot the left- and right-hand sides of the equation as separate functions and determine the intersection points. Note that, to ensure compatibility of the $x$x-axis values, radians must be used when plotting both functions.

Do: The display looks something like the following:

We can see that there are four solutions between $x=0$x=0 and $x=2\pi$x=2π.

Using technology, we obtain these approximate intersection points:

$\left(0.009,0.018\right)$(0.009,0.018)

$\left(1.53,0.084\right)$(1.53,0.084)

$\left(3.45,0.576\right)$(3.45,0.576)

$\left(3.99,0.992\right)$(3.99,0.992)

The solutions to the equation $e^{x-4}=\sin(2x)$ex−4=sin(2x) are the $x$x-coordinates of these points of intersection.

Practice questions

Question 3

Question 4

Question 5