Trigonometric equations are equations containing terms such as sin x, cos x and tan x.
They can be solved using the trigonometric graphs and, if necessary, a calculator. Another method using general solutions can be used.
Because trigonometric functions are periodic and continue forever, these trigonometric equations often have an infinite number of solutions unless the domain (xvalues) is fixed. Usually the domain will be given.
To illustrate the various methods that can be used, several different types of examples will be given. The solutions are given in the same units the question is written in. e.g. Degrees or radians.
The angles used in the special triangles often occur in trigonometric equations and are shown again below as a reminder.
Special Triangles
Angles such as 30° ( ), 45° ( ) and 60° ( ) are used frequently and the trigonometric ratios of these angles are obtained from two special triangles (see Unit 38, Year 12). They are summarised below:
sin 30°

cos 30°

tan 30°

sin 45°

cos 45°

tan 45°

sin 60°

cos 60°

tan 60°

1

√3

If the answers can be given using exact values from the special triangles these shoud be given. A calculator should only be used if special triangle angles are not involved.
Trigonometric Equations
Example 1
Solve sin x = 0.5, for 0° ≤ x ≤ 360°. Give answers in degrees.
Consider the functions y = sin x and the line y = 0.5. Where the line and curve meet will be the solutions. A calculator can be used to find the first value, by finding sin ^{1}(0.5)
A caclulator can be used for the first solution 30° and the second solution found from the symmetry of the graph (180°  30° = 150°).
The solution set is {30°, 150°}
Similar methods can be used for equations involving cosine and tangent.
Example 2
Solve 2sin^{2} x + sin x = 0 for 0 ≤ x ≤ 2π . Give answers in radians.
This is a quadratic equation, so factorise if possible.
sin x(2sin x + 1) = 0
There are two sets of solutions:
sin x = 0 and 2sin x + 1 = 0 which gives sin x = 0.5The solutions of sin x = 0 are 0, 3.14π and 2π
The solutions of sin x = 0.5 are 7π/6 and 11π/6
The solution set is {0, π, 7π/6, 11π/6, 2π}
Example 3
Solve √ 2cos 2x = 1 for 0 ≤ x ≤ 2π . Give answers in terms of π .
The cosine function is isolated by dividing both sides by √2. Because cos 2x is required the graph of cos x from 0 to 4π will need to be examined to find all of the roots.
√ 2 cos 2x = 1cos 2x = 1 / √ 2
The first solution can be found using the special triangles above or from a calculator. The other solutions are found from the symmetry of the graph:
2x = ,

2x = 2π − =

2x = 2π + =

2x = 4π − =

x =

x =

x =

x =

The solution set is {, ,, }
Example 4
Solve sin 3x + sin x = 0 for 0 ≤ x ≤ 2π . Give answers in terms of π .
The sum to product formula is used here.
2sin 2x cos x = 0
Therefore sin 2x = 0 or cos x = 0
2x = {0, π, 2π, 3π, 4π} or x = { , }