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Chapter 6
Trigonometric
Identities and
Equations
© 2011 Pearson Education, Inc.
All rights reserved
© 2010
2011 Pearson Education, Inc. All rights reserved
1
SECTION 6.5
Trigonometric Equations I
OBJECTIVES
1
2
3
4
Solve trigonometric equations of the form
a sin (x – c) = k, a cos (x – c) = k, and
a tan (x – c) = k.
Solve trigonometric equations by using the
zero-product property.
Solve trigonometric equations that contain
more than one trigonometric function.
Solve trigonometric equations by squaring
both sides.
TRIGONOMETRIC EQUATIONS
A trigonometric equation is a conditional
equation that contains a trigonometric
function with a variable.
An identity is an equation that is true for all
values in the domain of the variable.
Solving a trigonometric equation means
finding its solution set.
© 2011 Pearson Education, Inc. All rights reserved
3
EXAMPLE 1
Solving a Trigonometric Equation
Find all solutions of each equation. Express all
solutions in radians.
2
a. sin x 
2
3
b. cos  
2
c. tan x   3
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4
EXAMPLE 1
Solution
Solving a Trigonometric Equation
2
a. sin x 
2
a. First find all solutions in [0, 2π).
We know
I and II.
and sin x > 0 in quadrants
QI and QII angles with reference angles of
are
and
.
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5
EXAMPLE 1
Solving a Trigonometric Equation
Solution continued
Since sin x has a period of 2π, all solutions of the
equation are given by
or
for any integer n.
© 2011 Pearson Education, Inc. All rights reserved
6
EXAMPLE 1
Solving a Trigonometric Equation
3
b. cos  
2
Solution
a. First find all solutions in [0, 2π).

3
We know cos 
and cos θ < 0 in
6
2
quadrants II and III.

QII and QIII angles with reference angles of
6

7


5

are     
and     
.
6
6
© 2011 Pearson Education, Inc. All rights reserved
6
6
7
EXAMPLE 1
Solving a Trigonometric Equation
Solution continued
Since cos θ has a period of 2π, all solutions of
the equation are given by
5
7

 2n or  
 2n
6
6
for any integer n.
© 2011 Pearson Education, Inc. All rights reserved
8
EXAMPLE 1
Solving a Trigonometric Equation
c. tan x   3
Solution
a. Because tan x has a period of π, first find all
solutions in [0, π).

We know tan  3 and tan x < 0 in
3
quadrant II.

The QII angle with a reference angle of is
3
 2
.
x   
3
3
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9
EXAMPLE 1
Solving a Trigonometric Equation
Solution continued
Since tan x has a period of π, all solutions of the
equation are given by
2
x
 n
3
for any integer n.
© 2011 Pearson Education, Inc. All rights reserved
10
EXAMPLE 3
Solving a Linear Trigonometric Equation
Find all solutions in the interval [0, 2π) of the


equation 2sin  x    1  2.
4

Solution
Replace x 

4
2sin   1  2
2sin   1
1
sin  
2
with θ in the given equation.

The reference angle is
6
 1
 .
6 2
In QI and QII, sin θ > 0.
because sin
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11
EXAMPLE 3
Solving a Linear Trigonometric Equation
Solution continued
5
or


6
6
 
 5
or x  
x 
4 6
4
6
 
5 
x 
x

6 4
6 4
10 3 13
2 3 5
x


x


12 12 12
12 12 12

5 13 

The solution set in the interval [0, 2π) is  ,
.
 12 12 
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12
EXAMPLE 6
Solving a Quadratic Trigonometric Equation
Find all solutions of the equation
2
2sin   5sin   2  0.
Express the solutions in radians.
Solution
Factor 2sin   5sin   2  0.
 2sin   1 sin   2   0
 2sin   1  0 or sin   2   0
1
sin   2
sin  
2

5
No solution

or  
6
6
2
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13
EXAMPLE 6
Solving a Quadratic Trigonometric Equation
Solution continued

5
are the only solutions
and  
So,  
6
6
in the interval [0, 2π).
Since sin has a period of 2π, the solutions are


6
 2n
5
or  
 2n
6
for any integer n.
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14
EXAMPLE 8
Solving a Trigonometric Equation by
Squaring
Find all the solutions in the interval [0, 2π) to
the equation 3 cos x  sin x  1.
Solution
Square both sides and use identities to convert
to an equation containing only sin x.
3 cos x  sin x  1.

3 cos x

2
  sin x  1
2
3 cos 2 x  sin 2 x  2 sin x  1
3 1  sin x   sin x  2 sin x  1
2
2
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15
EXAMPLE 8
Solving a Trigonometric Equation by
Squaring
Solution continued
3  3sin 2 x  sin 2 x  2sin x  1
2
4 sin x  2sin x  2  0
2sin 2 x  sin x  1  0
 2sin x  1 sin x  1  0
2sin x 1  0
1
sin x 
2
or
sin x 1  0
sin x  1

5
x
or
6
6
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3
x
2
16
EXAMPLE 8
Solving a Trigonometric Equation by
Squaring
Solution continued
Possible solutions are:
x

6
5
x
6
3
x
2
3 cos

6
5
3 cos
6
3
3 cos
2
?
 sin

1
6
?
5
 sin
1
6
?
3
 sin
1
2
3 3

2 2
3 3
 
2 2
00
  3 
The solution set in the interval [0, 2π) is  , .
6 2 
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17
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