6.07 Trigonometric equations

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 1

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°

Practice question

Question 1

 

 

Solving trigonometric equations algebraically

The method for solving trigonometric equations is roughly the same. After any algebraic manipulation, you might need to undertake, use the positive value on the right-hand side of the equation to find the size of the related acute angle. Then, draw a quadrant (ASTC) diagram to locate the angles you require to make the equation true. Write down your solutions, ensuring that they are in the domain specified in the question. 

Worked examples

Example 2

Solve $\tan\theta=\sqrt{3}$tanθ=√3 in the domain $0^\circ\le\theta\le360^\circ$0°≤θ≤360°. 

Think: Let's first find the relative acute angle that satisfies this equation, and then use this angle to find the remaining solutions in the given domain.

Do: Using our knowledge of exact value triangles, we have that:

$\tan\theta$tanθ	$=$=	$\sqrt{3}$√3	(Given)
$\theta$θ	$=$=	$60^\circ$60°	(Using our knowledge of exact value triangles)

 

So $\theta=60^\circ$θ=60° is our relative acute angle. We need $\tan\theta$tanθ to be positive, so our quadrant diagram below shows us that we need to find angles in the first and third quadrant:

Hence, the two solutions will be the acute relative angle we found, $60^\circ$60°, and the equivalent relative angle in the third quadrant, which is $180^\circ+60^\circ$180°+60°. So our solutions are:

$\theta$θ

$=$=

$60^\circ,240^\circ$60°,240°

Types of trigonometric equations

There are many variations on trigonometric equations you need to be comfortable with. In this set, we will consider five types of equations.

Type 1: equations involving exact values

If a question doesn't specify a need for your final answer to be rounded, this indicates that it involves an exact value solution!

Practice question

Question 2

Question 3

Type 2: equations involving non-exact values

Make sure your answer is rounded in the way specified in the question. This could be to various levels of accuracy, such as nearest degree, nearest minute or a certain number of decimal places.

Practice question

Question 4

Type 3: equations involving boundary angles

Questions, where the value specified, is $0$0 or $\pm1$±1 land on what we can refer to as boundary angles. It might be easier for you to solve these questions by considering the sketch of the trigonometric function in question, rather than trying to use your calculator or a quadrant diagram. The exception to this is when $\tan x$tanx equals to $\pm1$±1, where it's more valuable to consider the exact value triangles to find a solution.

Practice question

Question 5

Type 4: equations with a different domain

It doesn't matter how you arrive at your final solutions here, as long as you get there in the end! Many students prefer to solve an equation as if the domain was $0^\circ\le x\le360^\circ$0°≤x≤360° and then "move" these answers into the correct domain by adding or subtracting $360^\circ$360° until they find the solutions required. You might feel comfortable looking at your quadrant diagram after finding the relative acute angle and going straight to writing the solutions in the required domain.

Having a variety of domains might mean you have more or less than two solutions, which is typically what we expect when solving for $0^\circ\le x\le360^\circ$0°≤x≤360°.

Practice question

Question 6