Chapter 5 Analytic Trigonometry

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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 ( angles.

sides.

x

−

c

) =

k

Solve trigonometric equations involving multiple 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 © 2010 Pearson Education, Inc. All rights reserved

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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.

a. sin

x

 2 2 b. cos    2 3 c. tan

x

  3 © 2010 Pearson Education, Inc. All rights reserved

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EXAMPLE 1 Solving a Trigonometric Equation a. sin

x

 2 Solution 2 a. First find all solutions in [0, 2

π

). We know and sin

x

> 0 only in quadrants I and II. QI and QII angles with reference angles of are: and © 2010 Pearson Education, Inc. All rights reserved

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EXAMPLE 1 Solving a Trigonometric Equation Solution continued Since sin

x

has a period of 2

π

, all solutions of the equation are given by or for any integer

n

.

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EXAMPLE 1 Solving a Trigonometric Equation b. cos    3 Solution 2 a. First find all solutions in [0, 2

π

).  We know and cos 6 2 quadrants II and III. 3

x

< 0 only in QII and QIII angles with reference angles of    6 5 6    © 2010 Pearson Education, Inc. All rights reserved  6 7 6   6

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EXAMPLE 1 Solving a Trigonometric Equation Solution continued Since cos

x

has a period of 2

π

, all solutions of the equation are given by   5  6  2

n

 or   7  6  2

n

 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

). quadrant II.

tan  3  3

x

< 0 only in 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  3 

n

 for any integer

n

.

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EXAMPLE 3 Solving a Linear Trigonometric Equation Find all solutions in the interval [0, 2

π

) of the equation: 2sin 

x

  4 2 Solution Replace 2sin   

x

  1  2  4   2sin  sin    1 1 2 by  in the given equation.

We know sin   1 6 2 sin    > 0 in Q I and II  6 ,   5 6  © 2010 Pearson Education, Inc. All rights reserved

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EXAMPLE 3 Solving a Linear Trigonometric Equation

x x x

     12   4  6 2    6  6  4 3  12  5  12 or Solution set in [0, 2

π

) is

x x x

5     12 ,   4 5    6  5  5 6  6  4 10   3  12 13  12 .

12 © 2010 Pearson Education, Inc. All rights reserved  13  12

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Solving a Trigonometric Equation EXAMPLE 4 Containing Multiple Angles 1 Find all solutions of the equation in 2 the interval [0, 2

π

).

Solution Recall cos  3  so 1 2  .

The period of cos

x

So 3

x

  3  2

n

 cos   ,   5  3 3 is 2

π

. Replace  or  > 0 in Q I and IV, 3

x

 with 3

x

.

5 3   2

n

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EXAMPLE 4 Solving a Trigonometric Equation Containing Multiple Angles Solution continued Or

x

  9  2

n

3  or

x

 5 9   2

n

To find solutions in the interval [0, 2

π

), try: 3 

n

= –1

n

= 0

n

= 1

x x x

    9   9 9   2 3 3  2     7 9  5  9

x



x



x

 © 2010 Pearson Education, Inc. All rights reserved 5  9 5  9 5  9  2  3  2  3    9  11  9

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EXAMPLE 4 Solving a Trigonometric Equation Containing Multiple Angles

n n

Solution continued = 2

x

  9  4 3  = 3

x

  9  2    13 9 19  9 

x



x

 5 9 9  5    4  2 3    17 9 23   9 Values resulting from

n

= –1 are too small.

Values resulting from

n

= 3 are too large.

Solutions we want correspond to

n

Solution set is    9 , 5  9 , 7 9  , 11 9  = 0, 1, and 2.

, 13 9  , 17 9    .

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EXAMPLE 7 Solving a Quadratic Trigonometric Equation Find all solutions of the equation 2sin 2   5sin   2  0.

Express the solutions in radians.

Solution Factor  2sin   2sin 2 2sin     5sin      2 2      0.

 0 2 sin   1 2      6 or   2 0  5 6  No solution because –1 ≤ sin © 2010 Pearson Education, Inc. All rights reserved  ≤ 1.

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EXAMPLE 7 Solving a Quadratic Trigonometric Equation Solution continued So,    6 and   5  6 are the only two solutions in the interval [0, 2

π

).

Since sin  has a period of 2

π

, the solutions are    6  2

n

 or   5  6  2

n

 , for any integer

n

.

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EXAMPLE 8 Solving a Trigonometric Equation Using Identities Find all the solutions of the equation 2sin 2   3 cos  0 in the interval [0, 2π).

Solution Use the Pythagorean identity to rewrite the equation in terms of cosine only.

2  2 si n 2   s 2     3 cos  3 co s  0  1  0 2  2cos 2   3 cos  0 2   3 cos   0 © 2010 Pearson Education, Inc. All rights reserved

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EXAMPLE 8 Solving a Trigonometric Equation Using Identities Solution continued 2 cos 2   3 cos   3  0 Use the quadratic formula to solve this equation.

2  cos   cos   3  4 3  24  3  4 27  3  3 3 4 © 2010 Pearson Education, Inc. All rights reserved

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EXAMPLE 8 Solving a Trigonometric Equation Using Identities Solution continued So, cos  cos  cos     3  3 3 4 3 4 4 3  1 or cos  cos  cos    3  3 3

 

4   4 3 2 3 No solution because –1 ≤ cos  ≤ 1.

cos  < 0 in QII, QIII © 2010 Pearson Education, Inc. All rights reserved

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EXAMPLE 8 Solving a Trigonometric Equation Using Identities Solution continued cos     2 3 when     6  5  6   6      6  7  6 Solution set in the interval [0, 2

π

) is 5  6 , 7  6   .

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Solving a Trigonometric Equation by EXAMPLE 9 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 3 cos

x



2

x

  sin

x



sin

x

 1.

 1



2 3  1  3 si n 2 2

x x

  sin 2  sin 2

x

 s n

x

 1

x



x

 1 © 2010 Pearson Education, Inc. All rights reserved

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EXAMPLE 9 Solving a Trigonometric Equation by Squaring Solution continued  4 sin 2

x

 3  3sin 2sin

x

 2

x

2   sin 2 0  2sin 2 2sin

x



x

  sin

x x

 1   0 0

x

 2sin

x

 1 2sin

x

 1  0 sin

x

 1

x

 2  6 or sin

x

 1  0 or 5  6 © 2010 Pearson Education, Inc. All rights reserved sin

x x

  1  3  2

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EXAMPLE 9 Solving a Trigonometric Equation by Squaring Solution continued Possible solutions are:

x x x

    6 5 6  3  2  3 cos 6 5  3 cos 6 3  3 cos 2 ?

 sin ?

 sin ?

 sin  6 5    6 1 1 3   1 2 Solution set in the interval [0, 2

π

) is 3 2  3 2  3 2  3 2 0  0  6 , 3  2   .

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