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5-3 Solving Trigonometric Equations
Solve each equation for all values of x.
1. 5 sin x + 2 = sin x
SOLUTION: The period of sine is 2π, so you only need to find solutions on the interval
and . Solutions on the interval (–
general form of the solutions is
,
. The solutions on this interval are
), are found by adding integer multiples of 2π. Therefore, the
+ 2nπ,
+ 2nπ,
.
3. 2 = 4 cos2 x + 1
SOLUTION: The period of cosine is 2π, so you only need to find solutions on the interval
are
,
,
, and
. Solutions on the interval (–
Therefore, the general form of the solutions is
,
+ 2nπ,
. The solutions on this interval
), are found by adding integer multiples of 2π.
+ 2nπ,
+ 2nπ,
+ 2nπ,
.
5. 9 + cot2 x = 12
SOLUTION: The period of cotangent is π, so you only need to find solutions on the interval
are
and
. Solutions on the interval (–
general form of the solutions is
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7. 3 csc x = 2 csc x +
+ nπ,
,
. The solutions on this interval
), are found by adding integer multiples of π. Therefore, the
+ nπ,
.
Page 1
are
,
,
, and
. Solutions on the interval (–
5-3 Therefore,
Solving Trigonometric
Equations
the general form of the
solutions is
,
), are found by adding integer multiples of 2π.
+ 2nπ,
+ 2nπ,
+ 2nπ,
+ 2nπ,
.
5. 9 + cot2 x = 12
SOLUTION: The period of cotangent is π, so you only need to find solutions on the interval
are
and
. Solutions on the interval (–
general form of the solutions is
+ nπ,
. The solutions on this interval
), are found by adding integer multiples of π. Therefore, the
,
+ nπ,
.
7. 3 csc x = 2 csc x +
SOLUTION: The period of cosecant is 2π, so you only need to find solutions on the interval
are
and
. Solutions on the interval (–
general form of the solutions is
+ 2nπ,
,
. The solutions on this interval
), are found by adding integer multiples of 2π. Therefore, the
+ 2nπ,
.
9. 6 tan2 x – 2 = 4
SOLUTION: The period of tangent is π, so you only need to find solutions on the interval
and
. Solutions on the interval (–
form of the solutions is
11. 7 cot x – + nπ,
+ nπ,
,
. The solutions on this interval are
), are found by adding integer multiples of π. Therefore, the general
.
= 4 cot x
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The period of cotangent is π, so you only need to find solutions on the interval
Page 2
. The only solution on this
and
. Solutions on the interval (–
), are found by adding integer multiples of π. Therefore, the general
,
5-3 form
Solving
of theTrigonometric
solutions is + nπ, Equations
+ nπ,
11. 7 cot x – .
= 4 cot x
SOLUTION: The period of cotangent is π, so you only need to find solutions on the interval
interval is
. Solutions on the interval (–
form of the solutions is
+ nπ,
,
. The only solution on this
), are found by adding integer multiples of π. Therefore, the general
.
Find all solutions of each equation on [0, 2 ).
13. sin4 x + 2 sin2 x − 3 = 0
SOLUTION: when x =
On the interval [0, 2π),
and when x =
. Since
is not a real number, the yields no additional solutions.
equation
15. 4 cot x = cot x sin2 x
SOLUTION: The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x = 0 has
solutions
and .
2
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17. cos3Manual
x + cos
x – cos
x=1
SOLUTION: Page 3
The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x = 0 has
5-3 solutions
Solving Trigonometric
Equations
and .
17. cos3 x + cos2 x – cos x = 1
SOLUTION: On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of π.
19. TENNIS A tennis ball leaves a racket and heads toward a net 40 feet away. The height of the net is the same
height as the initial height of the tennis ball.
a. If the ball is hit at 50 feet per second, neglecting air resistance, use d =
2
v0 sin 2
to find the interval of
possible angles of the ball needed to clear the net.
b. Find if the initial velocity remained the same but the distance to the net was 50 feet.
SOLUTION: a.
The interval is [15.4°, 74.6°].
b.
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Page 4
5-3 Solving Trigonometric Equations
On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of π.
19. TENNIS A tennis ball leaves a racket and heads toward a net 40 feet away. The height of the net is the same
height as the initial height of the tennis ball.
a. If the ball is hit at 50 feet per second, neglecting air resistance, use d =
2
v0 sin 2
to find the interval of
possible angles of the ball needed to clear the net.
b. Find if the initial velocity remained the same but the distance to the net was 50 feet.
SOLUTION: a.
The interval is [15.4°, 74.6°].
b.
If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.
Find all solutions of each equation on the interval [0, 2 ).
21. 1 = cot 2 x + csc x
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SOLUTION: Page 5
5-3 Solving Trigonometric Equations
If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°.
Find all solutions of each equation on the interval [0, 2 ).
21. 1 = cot 2 x + csc x
SOLUTION: Therefore, on the interval [0, 2π) the solutions are
,
, and
.
23. tan 2 x = 1 – sec x
SOLUTION: Therefore, on the interval [0, 2π) the solutions are 0,
, and
.
25. 2 – 2 cos2 x = sin x + 1
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Page 6
5-3 Therefore,
Solving on
Trigonometric
the interval [0, 2π)Equations
the solutions are 0,
, and
.
25. 2 – 2 cos2 x = sin x + 1
SOLUTION: Therefore, on the interval [0, 2π) the solutions are
,
, and
.
27. 3 sin x = 3 – 3 cos x
SOLUTION: Therefore, on the interval [0, 2π) the only valid solutions are
29. sec2 x – 1 + tan x –
and 0.
tan x =
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5-3 Therefore,
Solving on
Trigonometric
the interval [0, 2π)Equations
the only valid solutions are
29. sec2 x – 1 + tan x –
and 0.
tan x =
SOLUTION: Therefore, on the interval [0, 2π) the solutions are
,
,
and
.
31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive
power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical
components, PH and PV.
Using the equations above, determine for what values of
PV and PH are equivalent.
SOLUTION: The sine and cosine have the same values in the interval [0, 2π) at
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Therefore,
components
will
be equivalent when
Find all solutions of each equation on the interval [0, 2 ).
and .
.
Page 8
5-3 Therefore,
Solving on
Trigonometric
the interval [0, 2π)Equations
the solutions are
,
,
and
.
31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive
power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical
components, PH and PV.
Using the equations above, determine for what values of
PV and PH are equivalent.
SOLUTION: and The sine and cosine have the same values in the interval [0, 2π) at
Therefore, the components will be equivalent when
.
.
Find all solutions of each equation on the interval [0, 2 ).
33. +
= –4
SOLUTION: On the interval [0, 2π),cos x = –
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35. cot x cos x + 1 =
+ when x =
and when x =
.
Page 9
and The sine and cosine have the same values in the interval [0, 2π) at
5-3 Therefore,
Solving the
Trigonometric
Equations
components will be
equivalent when
.
.
Find all solutions of each equation on the interval [0, 2 ).
33. +
= –4
SOLUTION: On the interval [0, 2π),cos x = –
35. cot x cos x + 1 =
when x =
and when x =
.
+ SOLUTION: On
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,
Page 10
the interval
[0, 2π),cos x = – Equations
when x =
5-3 On
Solving
Trigonometric
35. cot x cos x + 1 =
and when x =
.
+ SOLUTION: On
when x =
and when x =
,
.
GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the
nearest hundredth.
37. sin x + cos x = 3x
SOLUTION: On the interval
, the only solution is when x = 0.41.
39. x log x + 5x cos x = –2
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Page 11
5-3 On
Solving
Trigonometric
the interval
, the only Equations
solution is when x = 0.41.
39. x log x + 5x cos x = –2
SOLUTION: On the interval
, the solutions are when x = 1.84 and when x = 4.49.
Find the x-intercepts of each graph on the interval [0, 2 ).
41. SOLUTION: Let y = 0 and solve for x.
On the interval [0, 2π) cos x = 0 when x =
43. eSolutions Manual - Powered by Cognero
SOLUTION: Let y = 0 and solve for x.
and x =
.
Page 12
the interval
[0, 2π) cos x = 0Equations
when x = and x =
5-3 On
Solving
Trigonometric
.
43. SOLUTION: Let y = 0 and solve for x.
and x =
On the interval [0, 2π) cot x = 1 when x =
.
Find all solutions of each equation on the interval [0, 4 ).
45. 4 tan x = 2 sec2 x
SOLUTION: On the interval [0, 4π) the solutions are
,
,
, and
.
47. csc x cot2 x = csc x
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Page 13
the interval
[0, 4π) the solutions
are ,
5-3 On
Solving
Trigonometric
Equations
,
, and
,
,
.
47. csc x cot2 x = csc x
SOLUTION: On the interval [0, 4π) the solutions are
,
,
,
,
, and
.
49. GEOMETRY Consider the circle below.
a. The length s of
is given by s = r(2 ) where 0 ≤ ≤ . When s = 18 and AB = 14, the radius is r =
Use a graphing calculator to find the measure of 2
.
in radians.
b. The area of the shaded region is given by A =
. Use a graphing calculator to find the radian measure
of θ if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth.
SOLUTION: a. Rewrite the arclength formula using s = 18 and r =
.
On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.
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The value of 2θ = 2(1.1968) or about 2.39 radians.
Page 14
5-3 Solving Trigonometric Equations
On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ.
The value of 2θ = 2(1.1968) or about 2.39 radians.
b. First, substitute into the given area formula and rearrange it.
Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ.
When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.
Solve each inequality on the interval [0, 2 ).
51. 0 < 2 cos x –
SOLUTION: Graph y = 2 cos x –
< 2 cos x –
.
on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0
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The zeros of y = 2 cos x –
Page 15
are about 0.785 or
and about 5.498 or . Therefore, 0 < 2 cos x –
on 0 ≤ 5-3 Solving Trigonometric Equations
When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians.
Solve each inequality on the interval [0, 2 ).
51. 0 < 2 cos x –
SOLUTION: on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0
Graph y = 2 cos x –
< 2 cos x –
.
The zeros of y = 2 cos x –
x<
or 53. are about 0.785 or
and about 5.498 or . Therefore, 0 < 2 cos x –
on 0 ≤ < x < 2π.
≤ tan x cot x
SOLUTION: Graph y =
and y = tan x cot x on [0, 2π). Use the intersect feature under the CALC menu to determine
on what interval(s)
≤ tan x cot x.
The graphs intersect at about 3.142 or π. Therefore,
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55. sin x − 1 < 0
SOLUTION: ≤ tan x cot x on 0 ≤ x < 2π.
Page 16
The zeros of y = 2 cos x –
are about 0.785 or
and about 5.498 or . Therefore, 0 < 2 cos x –
on 0 ≤ < or Trigonometric
< x < 2π.
5-3 xSolving
Equations
53. ≤ tan x cot x
SOLUTION: and y = tan x cot x on [0, 2π). Use the intersect feature under the CALC menu to determine
Graph y =
≤ tan x cot x.
on what interval(s)
The graphs intersect at about 3.142 or π. Therefore,
55. ≤ tan x cot x on 0 ≤ x < 2π.
sin x − 1 < 0
SOLUTION: sin x − 1. Use the zero feature under the CALC menu to determine on what interval(s)
Graph y =
1 < 0.
The zeros of
or sin x − 1 are about 0.785 or
and about 2.356 or + n , x =
+ n , x =
+ n , and x =
= tan x. Vijay thinks that the solutions
+ n . Alicia thinks that the solutions are x =
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and x =
sin x − 1 < 0 on 0 ≤ x <
< x < 2π.
57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +
are x =
. Therefore,
sin x −
+ n . Is either of them correct? Explain your reasoning.
+ n
Page 17
sin x − 1 are about 0.785 or
The zeros of
and about 2.356 or sin x − 1 < 0 on 0 ≤ x <
. Therefore,
5-3 Solving
Trigonometric
Equations
< 2π.
or < x
57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x +
are x =
+ n , x =
+ n , x =
+ n , and x =
and x =
+ n . Is either of them correct? Explain your reasoning.
= tan x. Vijay thinks that the solutions
+ n . Alicia thinks that the solutions are x =
+ n
SOLUTION: 2
First, solve tan x – tan x +
= tan x.
and x =
On [0, 2π) tan x = 1 when x =
and tan x =
when x =
and x =
.
Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest form. For
+ n
example, his solutions of x =
1,
+ nπ is equivalent to
and x =
+ n
could simply be stated as x =
+ n
because when n =
.
CHALLENGE Solve each equation for all values of x.
59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3
SOLUTION: On [0, 2 ) sin x = 1 when x =
when x =
and x =
, sin x = –1 when x =
, sin x = –
and x =
when x =
, and sin x =
.
Therefore, after checking for extraneous solutions, the solutions are
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,
,
,
,
, and
OPEN ENDED Write a trigonometric equation that has each of the following solutions.
.
Page 18
Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest form. For
+ n
example, his solutions of x =
and x =
+ n
could simply be stated as x =
+ n
because when n =
5-3 1,
Solving
Trigonometric
equivalent to
.Equations
+ nπ is
CHALLENGE Solve each equation for all values of x.
59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3
SOLUTION: On [0, 2 ) sin x = 1 when x =
when x =
and x =
, sin x = –1 when x =
, sin x = –
and x =
when x =
, and sin x =
.
Therefore, after checking for extraneous solutions, the solutions are
,
,
,
,
, and
.
OPEN ENDED Write a trigonometric equation that has each of the following solutions.
61. SOLUTION: Sample answer: When sin x = 0, x = 0 and x = π. When sin x =
solutions of 0, π,
, and
is 2
= 0 or sin x –
,x=
and x =
. So, one equation that has
2
= 0. This can be rewritten as 2 sin x =
sin x.
63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying
identities.
SOLUTION: Sample answer: When solving an equation, you use properties of equality to manipulate each side of the equation to
isolate a variable. When verifying an identity, you transform an expression on one side of the identity into the
expression on the other side through a series of algebraic steps.
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Verify each identity.
65. =
Page 19
solutions of 0, π,
, and
is 2
= 0 or sin x –
2
= 0. This can be rewritten as 2 sin x =
5-3 Solving
sin x. Trigonometric Equations
63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying
identities.
SOLUTION: Sample answer: When solving an equation, you use properties of equality to manipulate each side of the equation to
isolate a variable. When verifying an identity, you transform an expression on one side of the identity into the
expression on the other side through a series of algebraic steps.
Verify each identity.
65. =
SOLUTION: Find the value of each expression using the given information.
67. tan
; sin θ =
, tan
> 0
SOLUTION: Use the Pythagorean Identity that involves sin θ.
Since tan
is positive and sin is positive, cos θ must be positive. So,
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.
Page 20
5-3 Solving Trigonometric Equations
Find the value of each expression using the given information.
67. tan
; sin θ =
, tan
> 0
SOLUTION: Use the Pythagorean Identity that involves sin θ.
Since tan
69. sec
.; tan
is positive and sin is positive, cos θ must be positive. So,
= –1, sin
.
< 0
SOLUTION: Use the Pythagorean Identity that involves tan
Since tan
.
is negative and sin θ is negative, cos
2
Given
f (x)- Powered
= 2x –by5x
+ 3 and g(x)
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Cognero
71. (f – g)(x)
SOLUTION: and
= 6x + 4, find each.
or sec θ must be positive. Therefore,
.
Page 21
5-3 Since
Solving
Equations
tan Trigonometric
negative, cos
is negative and sin θ is
and
or sec θ must be positive. Therefore,
.
2
Given f (x) = 2x – 5x + 3 and g(x) = 6x + 4, find each.
71. (f – g)(x)
SOLUTION: 73. (x)
SOLUTION: 75. SAT/ACT For all positive values of m and n, if
= 2, then x =
A
B
C
D
E
SOLUTION: The correct answer is D.
77. Which
of the
following
is not
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A
a solution of 0 = sin
2
+ cos tan
?
Page 22
5-3 Solving Trigonometric Equations
The correct answer is D.
77. Which of the following is not a solution of 0 = sin
2
+ cos tan
?
A
B
C 2π
D
SOLUTION: Try choice A.
Try choice B.
Try choice C.
Try choice D.
The correct answer is D.
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