Download Section 5.2 - University of South Florida

Chapter 5
Analytic
Trigonometry
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All rights reserved
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1
SECTION 5.2
Trigonometric Equations
OBJECTIVES
1
2
3
4
5
Solve trigonometric equations of the form
a sin (x − c) = k, a cos (x − c) = k, and
a sin (x − c) = k
Solve trigonometric equations involving multiple
angles.
Solve trigonometric equations by using the zeroproduct property.
Solve trigonometric equations that contain more
than one trigonometric function.
Solve trigonometric equations by squaring both
sides.
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2
TRIGONOMETRIC EQUATIONS
A trigonometric equation is an equation that
contains a trigonometric function with a
variable.
Equations that are true for all values in the
domain of the variable are called identities.
Solving a trigonometric equation means to
find its solution set.
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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
Solving a Trigonometric Equation
2
a. sin x 
Solution
2
a. First find all solutions in [0, 2π).
We know
and sin x > 0 only in
quadrants I and II.
QI and QII angles with reference angles of
are:
and
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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.
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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 x < 0 only in
6
2
quadrants II and III.

QII and QIII angles with reference angles of
6
 7
 5
are:     
and     
6
6
6
6
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EXAMPLE 1
Solving a Trigonometric Equation
Solution continued
Since cos x has a period of 2π, all solutions of
the equation are given by
5
7

 2n or  
 2n
6
6
for any integer n.
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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 only in
3
quadrant II.

The QII angle with a reference angle of
is:
3
 2
   
3
3
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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

 n
3
for any integer n.
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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   by  in the given equation.

4
 1
2sin   1  2
We know sin 
6 2
2sin   1
sin > 0 in Q I and II
1

5
sin  
 , 
2
6
6
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EXAMPLE 3
Solving a Linear Trigonometric Equation
Solution continued
5
or


6
6
 5
 
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 
Solution set in [0, 2π) is  ,
.
 12 12 

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Solving a Trigonometric Equation
Containing Multiple Angles
EXAMPLE 4
1
Find all solutions of the equation cos 3x  in
2
the interval [0, 2π).
Solution

1
Recall cos  . cos > 0 in Q I and IV,
3 2

5
so   ,  
3
3
The period of cos x is 2π. Replace  with 3x.

5
3x   2n
3x 
 2n
So
or
3
3
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Solving a Trigonometric Equation
Containing Multiple Angles
EXAMPLE 4
Solution continued
2n
5 2n
x 
x

Or
or
9
3
9
3
To find solutions in the interval [0, 2π), try:

n = –1
n=0
n=1

2
5
x 

9
3
9
x

9

2 7
x 

9
3
9
5 2

x


9
3
9
5
x
9
5 2 11
x


9
3
9
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EXAMPLE 4
Solving a Trigonometric Equation
Containing Multiple Angles
Solution continued
n=2
n=3

4 13
x 

9
3
9

19
x   2 
9
9
5 4 17
x


9
3
9
5
23
x
 2 
9
9
Values resulting from n = –1 are too small.
Values resulting from n = 3 are too large.
Solutions we want correspond to n = 0, 1, and 2.
  5 7 11 13 17 
,
,
,
.
Solution set is  , ,
9
9 
9 9 9 9
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15
EXAMPLE 7
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
No solution because

5

or  
–1 ≤ sin ≤ 1.
6
6
2
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EXAMPLE 7
Solving a Quadratic Trigonometric Equation
Solution continued

5
are the only two
and  
So,  
6
6
solutions 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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17
EXAMPLE 8
Solving a Trigonometric Equation Using
Identities
Find all the solutions of the equation
2sin 2   3 cos  1  0 in the interval [0, 2π).
Solution
Use the Pythagorean identity to rewrite the
equation in terms of cosine only.
2 sin 2   3 cos  1  0
2 1  cos    3 cos  1  0
2
2  2cos 2   3 cos  1  0
3  2cos 2   3 cos  0
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Solving a Trigonometric Equation Using
Identities
EXAMPLE 8
Solution continued
2 cos   3 cos  3  0
2
Use the quadratic formula to solve this equation.
cos 
cos 

  3
  3 
2
 4  2  3
2  2
3  3  24

4
3  27

4
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33 3
4
19
EXAMPLE 8
Solving a Trigonometric Equation Using
Identities
Solution continued
So,
33 3
cos  
4
4 3
cos 
4
cos  3  1
No solution because
–1 ≤ cos ≤ 1.
or
33 3
cos 
4
2  3

cos  
4
3
cos   
2
cos < 0 in QII, QIII
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EXAMPLE 8
Solving a Trigonometric Equation Using
Identities
Solution continued
3

cos 
when  
2
6
 5
   
6
6

7
   
6
6
 5 7 
Solution set in the interval [0, 2π) is  ,  .
 6 6 
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21
Solving a Trigonometric Equation by
Squaring
EXAMPLE 9
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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EXAMPLE 9
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

5
x
or
6
6
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sin x  1
3
x
2
23
EXAMPLE 9
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 
Solution set in the interval [0, 2π) is  ,  .
6 2 
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