Section 7.5

Solving Trigonometric

Equations

Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

Objectives

Solve trigonometric equations.

Solve Trigonometric Equations

When an equation contains a trigonometric expression with a variable, such as cos x, it is called a trigonometric equation. Some trigonometric equations are identities, such as sin2 x + cos2 x = 1. Now we consider equations, such as 2 cos x = –1, that are usually not identities. As we have done for other types of equations, we will solve such equations by finding all values for x that make the equation true.

Example

Solve 2cos x 1.

First solve for cos x.

2π/3 and 4π/3 have cosine –1/2.

Solution:

23

2k or 43

2k

cos x 1

2

These numbers, plus any multiple of 2π are the solutions.

120ºk 360º or 240ºk 360º

Example

Solve 3tan2x 3 in the interval 0,2 .

First solve for tan 2x.

We are looking for solutions x to the equation for which

0 ≤ x < 2π.

Solution:tan2x 1

Multiplying by 2, we get 0 ≤ 2x < 4πwhich is the interval we use when solving tan 2x = –1.

Using the unit circle, find the points 2x in [0, 4π) for which tan 2x = –1.

They are:

Example (CONT)

The values of x, are found by dividing each of these by 2.

2x 34

,74

,11

4,15

4

x 38

,78

,11

8,15

8

Example

Solve 1

cos 1 1.2108 in [0º ,360º ).2

Solution: 1

2cos 1 1.2108

Use a calculator in DEGREE mode to find the reference angle = cos–1 0.4216 ≈ 65.06º.

1

2cos 0.2108

cos 0.4216

Example (cont)

Since cos is positive, the solutions are in quadrants I and IV. The solutions in [0º, 360º) are

65.06ºand360º – 65.06º = 294.94º

Example

Solve 2cos2 u 1 cosu in 0º ,360º .

Use the principal of zero products:

Solution: 2cos2 u 1 cosu

The solutions in [0º, 360º) are 60º, 180º and 300º.

2cos2 u cosu 1 0

2cosu 1 cosu 1 0

2cosu 1 0 or cosu 1 02cosu 1 or cosu 1

cosu 1

2 or cosu 1

u 60º ,300º or u 180º

Graphical Solution: INTERSECT METHOD

Example (cont)

and use the INTERSECT feature on the calculator.

y1 2cos2 x and y2 1 cos xGraph the equations:

The left most solution is 60º. Use the INTERSECT feature two more times to find the solutions, 180º and 300º.

Example (cont)

Graphical Solution: ZERO METHOD

2cos2 x cos x 1 0Write the equation in the form

The left most zero is 60º. Use the ZERO feature two more times to find the solutions, 180º and 300º.

Then graphy 2cos2 x cos x 1

Example

Solve 10sin2 x 12sin x 7 0 in 0º ,360º .

Use the quadratic formula: a = 10, b = –12, and c = –7.Solution:

No solution.

sin x 12 12 2 4 10 7

210

sin x 12 144 280

20

12 424

20

12 20.5913

20sin x 1.6296 or sin x 0.4296

reference angle: 25.44º

Example (cont)

Sin x is negative, the solutions are in quadrants III & IV.

The solutions are180º + 25.44º = 205.44ºand300º – 25.44º = 334.56º.

Example

Solve the following in [0, 2π).

Solution:

sin cos cotx x x

We cannot find the solutions algebraically. We can approximate them with a graphing calculator.

On the screen on the next slide, on the left side we use the INTERSECT METHOD. Graph

y1 sin x cos x and y2 cot x

On the screen on the right side we use the ZERO METHOD. Graph

y1 sin x cos x cot x

Example (cont)

The solutions in [0, 2π) are approximately 1.13 and 5.66.