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Chapter 5
Analytic
Trigonometry
5.5 Trigonometric
Equations
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
1
Objectives:
•
•
•
•
Find all solutions of a trigonometric equation.
Solve equations with multiple angles.
Solve trigonometric equations quadratic in form.
Use factoring to separate different functions in
trigonometric equations.
• Use identities to solve trigonometric equations.
• Use a calculator to solve trigonometric equations.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
2
Trigonometric Equations and Their Solutions
A trigonometric equation is an equation that contains a
trigonometric expression with a variable, such as sin x.
The values that satisfy such an equation are its solutions.
(There are trigonometric equations that have no solution.)
When an equation includes multiple angles, the period of
the function plays an important role in ensuring that we
do not leave out any solutions.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
3
Example: Finding all Solutions of a Trigonometric
Equation
Solve the equation: 5sin x  3sin x  3.
Step 1 Isolate the function on one side of the equation.
5sin x  3sin x  3
5sin x  3sin x  3sin x  3sin x  3
2sin x  3
3
sin x 
2
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
4
Example: Finding all Solutions of a Trigonometric
Equation
(continued)
Solve the equation: 5sin x  3sin x  3.
Step 2 Solve for the variable.
3
sin x 
2
Solutions for this equation in  0,2  are:
 2
,
3 3

2
The solutions for this equation are:  2n ,  2n
3
3
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
5
Example: Solving an Equation with a Multiple Angle
Solve the equation: tan 2 x  3,0  x  2 .
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
6
Example: Solving an Equation with a Multiple Angle
Solve the equation: tan 2 x  3,0  x  2 .
tan

3
 3
Because the period is  , all solutions for
this equation are given by

0



n0
x 

2 x   n
6 2 6
3
 n
x 
   3 4 2


n 1 x    
6 2
6 2 6 6
6
3
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
7
Example: Solving an Equation with a Multiple Angle
(continued)
Solve the equation: tan 2 x  3,0  x  2 .
Because the period is  , all solutions for this equation are
 n
given by x   .
6 2
n2
n3

2  6 7
x 
 

6 2 6 6
6
 3  9 10 5
x 
 


6 2 6 6
6
3

2 7
5
0,2

 , the solutions are: , , , and .
In the interval 
6 3 6
3
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
8
Example: Solving a Trigonometric Equation Quadratic in
Form
Solve the equation: 4cos x  3  0, 0  x  2 .
2
4cos 2 x  3  0
The solutions in the interval  0,2 
4cos x  3
2
for this equation are:
3
2
cos x 
4
 5 7
3
3
cos x  

4
2
11
, , , and
.
6 6 6
6
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
9
Example: Using Factoring to Separate Different
Functions
Solve the equation: sin x tan x  sin x, 0  x  2 .
sin x tan x  sin x
sin x tan x  sin x  0
tan x  1  0
sin x(tan x  1)  0
sin x  0
tan x  1
x

x 0 x 
4
5
x
4
The solutions for this equation in the interval  0,2  are:

5
0, ,  , and
.
4
4
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
10
Example: Using an Identity to Solve a Trigonometric
Equation
Solve the equation: cos2 x  sin x  0, 0  x  2 .
cos 2 x  sin x  0
1  2sin 2 x  sin x  0
The solutions in the
2sin 2 x  sin x  1  0
interval  0,2  are
(2sin x  1)(sin x  1)  0
p
7p
11p
2sin x  1  0
, , and
.
2 6
6
2sin x  1
sin x  1  0
1
sin x  
sin x  1
2
p
x=
7
11
x
x
2
6
6
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
11
Example: Solving Trigonometric Equations with a
Calculator
Solve the equation, correct to four decimal places, for
0  x  2 .
tanx is positive in quadrants I and III
tan x  3.1044
In quadrant I x  1.2592
1
x  tan (3.1044)
x  1.2592
In quadrant III x    1.2592
 4.4008
The solutions for this equation are 1.2592 and 4.4008.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
12
Example: Using a Calculator to Solve Trigonometric
Equations
Solve the equation, correct to four decimal places, for
0  x  2 .
Sin x is negative in quadrants III and IV
sin x  0.2315
In quadrant III x    0.2336
1
x  sin (0.2315)
x  3.3752
x  0.2336
In quadrant IV x  2  1.2592
x  6.0496
The solutions for this equation are 3.3752 and 6.0496.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
13