Download Section 7.5

CHAPTER 7:
Trigonometric Identities, Inverse
Functions, and Equations
7.1 Identities: Pythagorean and Sum and Difference
7.2 Identities: Cofunction, Double-Angle, and HalfAngle
7.3 Proving Trigonometric Identities
7.4 Inverses of the Trigonometric Functions
7.5 Solving Trigonometric Equations
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7.5
Solving Trigonometric Equations

Solve trigonometric equations.
Copyright © 2009 Pearson Education, Inc.
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.
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Slide 7.5 - 4
Example
Solve 2 cos x  1.
Solution:
First solve for cos x.
1
cos x  
2
2π/3 and 4π/3 have cosine –1/2.
These numbers, plus any
multiple of 2π are the solutions.
2
4
 2k or
 2k
3
3
120º k  360º or 240º k  360º
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Slide 7.5 - 5
Example
Solve 3tan 2x  3 in the interval 0,2 .
Solution:
First solve for tan 2x. tan2x  1
We are looking for solutions x to the equation for which
0 ≤ x < 2π.
Multiplying by 2, we get
0 ≤ 2x < 4π
which 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.
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Slide 7.5 - 6
Example
Solution continued:
They are:
3 7 11 15
2x 
, ,
,
4 4 4
4
The values of x, are found by
dividing each of these by 2.
3 7 11 15
x
, ,
,
8 8 8
8
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Slide 7.5 - 7
Example
1
Solve cos  1  1.2108 in [0º,360º ).
2
Solution:
1
cos   1  1.2108
2
1
cos   0.2108
2
cos  0.4216
Use a calculator in DEGREE mode to find the reference
angle = cos–1 0.4216 ≈ 65.06º.
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Slide 7.5 - 8
Example
Solution continued:
Since cos  is positive, the
solutions are in quadrants I
and IV. The solutions in
[0º, 360º) are
65.06º
and
360º – 65.06º = 294.94º
Copyright © 2009 Pearson Education, Inc.
Slide 7.5 - 9
Example
Solve 2 cos2 u  1 cosu in 0º, 360º .
Solution:
Use the principal
2cos2 u  1 cosu
of zero products:
2cos2 u  cosu 1  0
2 cosu  1cosu  1  0
2cosu 1  0 or cosu 1  0
2cosu  1 or cosu  1
1
cosu 
or cosu  1
2
u  60º, 300º or u  180º
The solutions in [0º, 360º) are 60º, 180º and 300º.
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Slide 7.5 - 10
Example
Solution continued:
Graphical Solution: INTERSECT METHOD
Graph the equations: y1  2 cos2 x
and
y2  1 cos x
and use the INTERSECT
feature on the calculator.
The left most solution is
60º. Use the INTERSECT
feature two more times to
find the solutions, 180º
and 300º.
Copyright © 2009 Pearson Education, Inc.
Slide 7.5 - 11
Example
Solution continued:
Graphical Solution: ZERO METHOD
2
Write the equation in the form 2cos x  cos x 1  0
Then graph
2
y  2 cos x  cos x  1
The left most zero is 60º.
Use the ZERO feature
two more times to find the
solutions, 180º and 300º.
Copyright © 2009 Pearson Education, Inc.
Slide 7.5 - 12
Example
Solve 10sin2 x 12sin x  7  0 in 0º, 360º .
Solution:
Use the quadratic formula: a = 10, b = –12, and c = –7.
sin x 
 12  
12 
2
 4 10 7 
2 10
12  144  280
12  424 12  20.5913

sin x 

20
20
20
sin x  1.6296 or sin x  0.4296
No solution.
reference angle: 25.44º
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Slide 7.5 - 13
Example
Solution continued:
Sin x is negative, the solutions are in quadrants III & IV.
The solutions are
180º + 25.44º = 205.44º
and
300º – 25.44º = 334.56º.
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Slide 7.5 - 14
Example
Solve each of the following in [0, 2π).
a) x 2 1.5  cos x
b) sin x  cos x  cot x
Solution:
We cannot find the solutions algebraically. We can
approximate them with a graphing calculator.
a. On the screen on the next slide, on the left side we
use the INTERSECT METHOD. Graph
y1  x 2  1.5 and y2  cos x
On the screen on the right side we use the ZERO
METHOD. Graph
y1  x 2  1.5  cos x
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Slide 7.5 - 15
Example
Solution continued:
The solution in [0, 2π) is approximately 1.32.
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Slide 7.5 - 16
Example
Solution continued:
b) sin x  cos x  cot 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
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Slide 7.5 - 17
Example
Solution continued:
The solutions in [0, 2π) are approximately 1.13 and 5.66.
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Slide 7.5 - 18