Lesson 9.2 Exercises, pages Exercises

Exercises

A

Use algebra to solve each equation. Give exact values when possible;

otherwise write the roots to the nearest degree or the nearest

hundredth of a radian. Verify the solutions.

4.

Solve each equation over the domain 0

… x 6 2p.

a) sin x

=

b) tan x

=

13

c) cos x

=

1

22

5.

Verify that each given value of x is a root of the equation.

a)

tan

_x

_{- 3 = 0; x =}

3

b) 8 sin

x

+ 6 sin x + 1 = 0; x =

B

6.

Solve each equation over the domain 0

… x 6 2p, then state the

general solution.

a)

3 cos x

- 2 = 0

b) 2 tan x

_{+ 15 = 0}

Lesson 9.2 Exercises, pages 637–642

The reference angle is: sin−1_a13 2 b = P 3 In Quadrant 1, x=P₃ In Quadrant 2, x= P −P₃, or 2₃P

The reference angle is: tan−1_a1 13b = P 6 In Quadrant 1, x=P₆ In Quadrant 3, x= P +P₆, or 7₆P

The reference angle is: cos−1_a1 12b = P 4 In Quadrant 1, x=P₄ In Quadrant 4, x= 2P −P₄, or 7₄P

Substitute x=P₃ in each side of the equation. L.S.= tan2_aP

3b− 3

= (13)2_{− 3} = 0

= R.S.

Since the left side is equal to the right side, the root is verified.

Substitute x= 7₆P in each side of the equation. L.S.= 8 sin2_a7P 6 b+ 6 sin 7 P 6 + 1 = 8 a−1₂b2+ 6 a−1₂b + 1 = 2 − 3 + 1 = 0 = R.S.

Since the left side is equal to the right side, the root is verified.

3 cos x= 2 cos x= 2₃ In Quadrant 1, x= cos−1_a 2 3b x= 0.8410. . . In Quadrant 4, x= 2P − 0.8410. . . x= 5.4421. . .

The roots are: x=

#

#

The general solution is: x=

#

#

2 tan x_{= −15} tan x= −1₂5

The reference angle is: tan−1_{a 1}5 2 b = 0.8410. . . In Quadrant 2, x= P − 0.8410. . . x= 2.3005. . . In Quadrant 4, x= 2P − 0.8410. . . x= 5.4421. . .

The roots are: x=

#

#

7.

Solve each equation for

-p … x … p.

a)

3 tan x

- 3 = 5 tan x - 1 b)

5(1

+ 2 sin x) = 2 sin x + 1

b) Solve each equation for

_{-90° … x … 180°.}

i)

_{4 sec x =}

_{-5 ii)}

_-

₌

csc x

iii)

5

13 cot x = 5

8.

a) Solve each equation for

-180° … x … 90°.

i)

2 csc x = 6 ii)

-6 = 3 cot x

iii)

12 sec x = 2

2 tan x= −2

tan x= −1

The reference angle is: tan−1₍₁₎₌P 4 In Quadrant 2, x= P −P₄ x= 3₄P In Quadrant 4, x= −P₄

The roots are: x= 3₄P and x= −P₄

5+ 10 sin x = 2 sin x + 1

8 sin x= −4

sin x= −1₂

The reference angle is: sin−1_a 1 2b = P 6 In Quadrant 3, x= −P +P₆ x= −5₆P In Quadrant 4, x= −P₆

The roots are: x= −P₆ and x= −5₆P

sec x= −5₄

cos x= −4₅

The reference angle is: cos−1_a 4

5b = 36.8698. . . ° In Quadrant 2,

csc x= −3₂

sin x= −2₃

The reference angle is: sin−1_a 2

3b = 41.8103. . . ° Quadrant 3 is not in the domain.

csc x= 3

sin x= 1₃

The reference angle is: sin−1_a 1

3b = 19.4712. . . ° In Quadrant 1, x= 19.4712. . . °

Quadrant 2 is not in the domain. The root is: x=

#

−2 = cot x

tan x= −1₂

The reference angle is: tan−1_a 1

2b = 26.5650. . . ° Quadrant 2 is not in the domain. In Quadrant 4, x= −26.5650. . . °

The root is: x=

#

sec x= 2 12

sec x= 12

cos x= 1 12

The reference angle is: cos−1_a 1

12b = 45°

In Quadrant 1, x= 45°

In Quadrant 4, x= −45°

The roots are: x= 45° and

x= −45°

cot x= 1 13

tan x= 13

The reference angle is: tan−1_{(13) = 60°} In Quadrant 1, x= 60°

Quadrant 3 is not in the domain.

9.

For each equation, determine the general solution over the set of

real numbers, then list the roots over the domain

-p … x 6 0.

a)

cos 3x

- 1 = 5 cos 3x + 2 b)

3 sin 4x = 3

- 2 sin 4x

10.

Two students determined the general solution of the equation

3 sin x

+ 5 = 5(sin x + 1). Joseph said the solution is x = 2pk or

x = p

+ 2pk, where k is an integer. Yeoun Sun said the solution

is x

= pk, where k is an integer. Who is correct? Explain.

RM, US, CR1

Both students are correct because both expressions produce the same roots: x= 0, x =tP, x = t2P, x = t3P, and so on

−3 = 4 cos 3x

cos 3x= −3₄

3x= cos−1_a−3 4b

The terminal arm of angle 3x lies in Quadrant 2 or 3. The reference angle for angle 3x is: cos−1_a 3 4b = 0.7227. . . In Quadrant 2, 3x= P − 0.7227. . . x= 0.8062. . . In Quadrant 3, 3x= P + 0.7227. . . x= 1.2881. . .

The period of cos 3x is 2₃P, so the general solution is: x=

#

#

In the given domain, the roots are: x= 0.8062. . . − 2₃P x=

#

#

#

The terminal arm of angle 4x lies in Quadrant 1 or 2. The reference angle for angle 4x is: sin−1_a 3 5b = 0.6435 . . . In Quadrant 1, 4x= 0.6435. . . x= 0.1608. . . In Quadrant 2, 4x= P − 0.6435. . . x= 0.6245. . .

The period of sin 4x is 2₄P, or P₂, so the general solution is: x=

#

#

In the given domain, the roots are: x= 0.1608. . . −P₂ x=

#

#

#

#

11.

Solve each equation over the domain

-p … x …

.

a)

4 cos

_x

_{- 3 = 0 b)}

_{2 tan}

_{x = 3}

12.

Use factoring to solve each equation over the domain

-90° … x 6 270°.

a)

2 cos x sin x

- cos x = 0 b)

3 tan x

+ tan

_{x = 2 tan x}

RM, US

13.

A student wrote the solution below to solve the equation

2 sin

_x

_{+ sin x = 1 over the domain 0 … x 6 2p.}

Identify any errors, then write a correct solution.

2 sin

_x

_{+ sin x = 1}

(sin x)(2 sin x

+ 1) = 1

sin x = 1 or 2 sin x

+ 1 = 1

x =

sin x = 0

The reference angle is: cos−1_a13 2 b = P 6 In Quadrant 1, x=P₆ In Quadrant 2, there is no solution in the given domain. In Quadrant 3, x= −P +P₆

x= −5₆P

In Quadrant 4, x= −P₆

The roots are: x=tP₆ and x= −5₆P

tan2_{x= 3}

2 tan x=t_A3₂

The reference angle is: tan−1_a

A

3

2 b = 0.8860. . . In Quadrant 1, x= 0.8860. . .

In Quadrant 2, there is no solution in the given domain.

In Quadrant 3, x= −P + 0.8860. . .

x= −2.2555. . .

In Quadrant 4, x= −0.8860. . .

The roots are: x=

#

x=

#

The roots are: x= 30°,

x=t90°, and x = 150° tan x+ tan2_{x= 0} (tan x)(1+ tan x) = 0 Either tan x= 0 x= 0° or x = 180° or 1+ tan x = 0 tan x= −1 x= 135° or x = −45°

The roots are: x= 0°, x = 180°,

x= 135°, and x = −45°

or sin x+ 1 = 0

14.

Solve each equation over the domain

-2p … x … 2p, then

determine the general solution.

a)

2 cos

_x

_{- cos x - 1 = 0 b)}

_{5 sin}

_x

_{+ 3 sin x = 2}

(2 cos x+ 1)(cos x − 1) = 0

For 2 cos x+ 1 = 0

2 cos x= −1

cos x= −1₂

The terminal arm of angle x lies in Quadrant 2 or 3.

The reference angle is: cos−1_a 1 2b = P 3 In Quadrant 2, x= P −P₃, or 2₃P and, x= −P −P₃, or −4₃P In Quadrant 3, x= P +P₃, or 4₃P and, x= −P +P₃, or −2₃P For cos x− 1 = 0 cos x= 1

The terminal arm lies along the positive x-axis.

So, x=t2P or x = 0

The roots are: x= 0, x =t2₃P, x=t4₃P, and x=t2P

Since consecutive roots differ by 2₃P, the general solution is: x= 2₃Pk, k { ℤ 5 sin2_{x+ 3 sin x − 2 = 0} (5 sin x− 2)(sin x + 1) = 0 For 5 sin x− 2 = 0 5 sin x= 2 sin x= 2₅

The terminal arm of angle x lies in Quadrant 1 or 2.

The reference angle is: sin−1_a 2 5b = 0.4115. . . In Quadrant 1, x= 0.4115. . . and, x= −2P + 0.4115. . . x= −5.8716. . . In Quadrant 2, x= P − 0.4115. . . x= 2.7300. . . and, x= −P − 0.4115. . . x= −3.5531. . . For sin x+ 1 = 0 sin x= −1

The terminal arm lies along the negative y-axis.

So, x= 3₂P or x= −P₂

The roots are: x=

#

x=

#

#

#

Since the period is 2P, the general solution is:

x=

#

#

C

15.

Write a second-degree trigonometric equation that has roots

,

,

over the domain 0

… x 6 2p.

RM, CR1, CR2

16.

Determine the number of roots each equation has over the domain

0

_{… x 6 2p, where 1 6 a 6 b.}

a)

(a cos x

_{- b)(b sin x + a) = 0}

b) (b sin

_x

_{- 1)(a tan x + b) = 0}

RM, US

Solve the equation. Either a cos x− b = 0

cos x_{= ba}

But b + a, so b_{a + 1, and there} is no real solution

Or b sin x+ a = 0

sin x= −a_b

Since b + a, there is a root in the two quadrants where the sine ratio is negative.

So, there are 2 roots.

Solve the equation. Either b sin2_{x− 1 = 0}

sin x=t_A1_b

Since b + 1, 0 * _A1_{b *}1, and there is a root in each quadrant. Or a tan x+ b = 0

tan x_{= −ba} There is a root in the two quadrants where the tangent ratio is negative.

So, there is a total of 6 roots. Sample response: Since sin P

6 = sin 5

P

6 , these two roots can be determined from one factor, so use the sine ratio. Work backward.

Either x=P₆ or x=P₂

So, sin x= 1₂ or sin x= 1

Then, sin x− 1₂= 0 or sin x − 1 = 0

So, an equation is: asin x −1₂b (sin x − 1) = 0

This can be written as: (2 sin x− 1)(sin x − 1) = 0 or