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Trigonometric Equations

Objectives
Upon completion of this lesson, you will be able to:

Solve trigonometric equations over the set of real numbers or on a given interval.

Required Reading
Swokowski/Cole: Section 6.2, pages 468-479.

Discussion

A trigonometric equation is simply an equation involving trigonometric functions. We apply the
same methods of solution (such as factoring) to trigonometric equations as we do to ordinary
equations.

Do not divide by any expression which could be zero.

Example 1: Solve sin2 x = sin x on the interval [0, 2π).

Solution:

It is tempting to divide both sides of the equation by the factor sin x .
If this is done, we have sin x = 1 and .
But in dividing by sin x , we divided by an unknown quantity that may be equal to 0 and
we lost valid solutions.

The correct solution is as follows :

sin2 x = sin x
sin2 x - sin x = 0 subtract sin x from both sides
sin x(sin x -1) = 0 factor
sin x = 0 or sin x -1= 0 set each factor equal to 0 and solve
sin x = 1

Hence x = 0,π,

Example 2: Solve 2cos2 x = 1+ sin x on the interval [0, 2π).

Solution: This equation will be quadratic in form once we put it in terms of one trigonometric
function. We know that cos2 x + sin2 x = 1, so cos2 x = 1- sin2 x .

2(1- sin2 x) = 1+ sin x substitute
2sin2 x + sin x -1 = 0 write in quadratic form
(2sin x -1)(sin x +1) = 0 factor
or sin x = -1
 
set each factor equal to 0 and solving for sin x

This gives us the following 3 solutions on the interval [0, 2π):

For trigonometric equations that do not factor, use the quadratic formula . The next
example demonstrates how to handle such a case.

Example 3:
Approximate to the nearest 0.1° the solutions on the interval [0°, 360°) for the
equation 2sin2θ - 5sinθ +1 = 0.
Solution: Quadratic Formula:

Since the equation is in quadratic form, ax2 + bx + c = 0, with x = sinθ, and a = 2, b = -5,
and c = 1, we have



sinθ = 2.2807764064 or sinθ = 0.219223593596

The equation sinθ = 2.2807764064 has no solutions, since -1< sinθ < 1 for all θ.

For sinθ = 0.219223593596, two solutions exist on the interval [0°, 360°], one in
quadrant I and the other in quadrant II where sinθ > 0.

To find the solutions, set your calculator to degree mode and θ = sin-1(0.219223593596) .

Hence, the reference angle is
The two solutions are:



Notes:

(1) Before factoring, be sure the expression is set equal to zero .
(2) Remember that -1< sinθ < 1 and -1 < cosθ < 1. This will help you recognize
extraneous values

Practice Problems
Work these problems. Answers to the odd numbered problems can be found at the end of your
text, even answers are below.

Section Pages Exercises
6.2 479-480 1, 3, 5, 7, 9, 11, 15, 19, 21, 23, 27, 31, 37, 39, 41, 42, 43, 45, 47, 49, 51,
61, 62, 63, 65, 66 (answer to the nearest 0.1° for 61, 62, 63, 65, 66)

Answers to even exercises.

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