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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
SOLUTION: eSolutions Manual - Powered by Cognero
The period of cotangent is π, so you only need to find solutions on the interval
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. 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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by Cognero
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 racquet 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 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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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 racquet 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 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 ).
2
21. 1 = Manual
cot x- +Powered
csc x by Cognero
eSolutions
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
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,
, and
.
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