Download 5-3 Solving Trigonometric Equations

interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
solutions is
5-3 Solving Trigonometric Equations
2nπ,
+ 2nπ,
+ 2nπ,
+ 2nπ,
+
.
3. 2 = 4 cos2 x + 1
Solve each equation for all values of x.
1. 5 sin x + 2 = sin x
SOLUTION: SOLUTION: The period of sine is 2π, so you only need to find
solutions on the interval
. The solutions on
and this interval are
. Solutions on the
interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
+ 2nπ,
solutions is
+ 2nπ,
.
The period of cosine is 2π, so you only need to find
solutions on the interval
. The solutions on
this interval are
2nπ,
SOLUTION: ,
, and
. Solutions on
the interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
solutions is
2. 5 = sec 2 x + 3
,
+ 2nπ,
+ 2nπ,
+ 2nπ,
+
.
4. 4 tan x – 7 = 3 tan x – 6
SOLUTION: The period of secant is 2π, so you only need to find
solutions on the interval
. The solutions on
The period of tangent is π, so you only need to find
solutions on the interval
. The only solution on
this interval are , , , and . Solutions on the
interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
this interval is
solutions is
Therefore, the general form of the solutions is
2nπ,
+ 2nπ,
+ 2nπ,
+ 2nπ,
+
.
3. 2 = 4 cos2 x + 1
SOLUTION: . Solutions on the interval (–
,
),
are found by adding integer multiples of π. nπ,
+
.
5. 9 + cot2 x = 12
SOLUTION: The period of cotangent is π, so you only need to find
solutions on the interval
. The solutions on this
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The period of cosine is 2π, so you only need to find
solutions on the interval
. The solutions on
interval are
,
and
. Solutions on the intervalPage
(– 1
), are found by adding integer multiples of π. are found by adding integer multiples of π. Therefore, the general form of the solutions is
+
5-3 Solving Trigonometric Equations
nπ,
interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
solutions is
.
5. 9 + cot2 x = 12
+ 2nπ,
+ 2nπ,
.
7. 3 csc x = 2 csc x +
SOLUTION: SOLUTION: The period of cosecant is 2π, so you only need to
find solutions on the interval
. The solutions
The period of cotangent is π, so you only need to find
solutions on the interval
. The solutions on this
interval are
,
and
. Solutions on the interval (–
), are found by adding integer multiples of π. Therefore, the general form of the solutions is
nπ,
+ nπ,
+
on this interval are
and
. Solutions on the
interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
solutions is
+ 2nπ,
+ 2nπ,
.
8. 11 = 3 csc2 x + 7
SOLUTION: .
6. 2 – 10 sec x = 4 – 9 sec x
SOLUTION: The period of secant is 2π, so you only need to find
solutions on the interval
. The solutions on
this interval are
and
. Solutions on the
interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
solutions is
+ 2nπ,
+ 2nπ,
.
The period of cosecant is 2π, so you only need to
find solutions on the interval
. The solutions
on this interval are
2nπ,
SOLUTION: ,
, and
. Solutions
on the interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
solutions is
7. 3 csc x = 2 csc x +
,
+ 2nπ,
+ 2nπ,
+ 2nπ,
+
.
9. 6 tan2 x – 2 = 4
SOLUTION: The period of cosecant is 2π, so you only need to
find solutions on the interval
. The solutions
on this interval are
and
. Solutions on the
interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
solutions is
+ 2nπ,
+ 2nπ,
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8. 11 = 3 csc2 x + 7
SOLUTION: .
The period of tangent is π, so you only need to find
solutions on the interval
. The solutions on this
interval are
,
and
Page 2
. Solutions on the interval (–
), are found by adding integer multiples of π. on the interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
solutions is
+ 2nπ,
+ 2nπ,
+ 2nπ,
+
5-3 Solving Trigonometric Equations
2nπ,
(– , ), are found by adding integer multiples of
2π. Therefore, the general form of the solutions is
+ 2nπ,
.
9. 6 tan2 x – 2 = 4
11. 7 cot x – + 2nπ,
.
= 4 cot x
SOLUTION: SOLUTION: The period of tangent is π, so you only need to find
solutions on the interval
. The solutions on this
The period of cotangent is π, so you only need to find
solutions on the interval
. The only solution on
interval are
this interval is
,
and
. Solutions on the interval (–
), are found by adding integer multiples of π. Therefore, the general form of the solutions is
nπ,
+ nπ,
+
. Solutions on the interval (–
,
are found by adding integer multiples of π. Therefore, the general form of the solutions is
nπ,
.
),
+
.
12. 7 cos x = 5 cos x +
10. 9 + sin2 x = 10
SOLUTION: SOLUTION: The period of cosine is 2π, so you only need to find
solutions on the interval
. The solutions on
The period of sine is 2π, so you only need to find
solutions on the interval
. The solutions on
this interval are
this interval are
and . Solutions on the interval
(– , ), are found by adding integer multiples of
2π. Therefore, the general form of the solutions is
+ 2nπ,
+ 2nπ,
and
. Solutions on the
interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
solutions is
+ 2nπ, + 2nπ,
.
.
Find all solutions of each equation on [0, 2 ).
11. 7 cot x – 13. sin4 x + 2 sin2 x − 3 = 0
= 4 cot x
SOLUTION: SOLUTION: The period of cotangent is π, so you only need to find
solutions on the interval
. The only solution on
eSolutions
by Cogneroon
this Manual
interval- Powered
is . Solutions
the interval (–
are found by adding integer multiples of π. ,
),
On the interval [0, 2π),
when x =
Page 3
and interval (– , ), are found by adding integer
multiples of 2π. Therefore, the general form of the
5-3 solutions
Solvingis Trigonometric
Equations
+ 2nπ, + 2nπ,
.
Find all solutions of each equation on [0, 2 ).
4
2
13. sin x + 2 sin x − 3 = 0
The equation cos x = 2 has no real solutions since
the maximum value the cosine function can obtain is
1. On the interval [0, 2π), the equation
has solutions 0 and π.
15. 4 cot x = cot x sin2 x
SOLUTION: SOLUTION: The equations sin x = 2 and sin x = –2 have no real
solutions. On the interval [0, 2π), the equation cot x =
when x =
On the interval [0, 2π),
when x =
. Since
and is not a real yields no number, the equation
additional solutions.
and 0 has solutions
.
16. csc2 x – csc x + 9 = 11
SOLUTION: 14. –2 sin x = –sin x cos x
SOLUTION: On the interval [0, 2π), the equation csc x = –1 has a
The equation cos x = 2 has no real solutions since
the maximum value the cosine function can obtain is
1. On the interval [0, 2π), the equation
has solutions 0 and π.
solution of
and the equation csc x = 2 has
solutions of and .
17. cos3 x + cos2 x – cos x = 1
SOLUTION: 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 .
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16. csc2 x – csc x + 9 = 11
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.
18. 2 sin2 x = sin x + 1
SOLUTION: Page 4
5-3
On the interval [0, 2π), the equation cos x = 1 has a
solution
of Trigonometric
0 and the equation cosEquations
x = –1 has a
Solving
solution of π.
18. 2 sin2 x = sin x + 1
The interval is [15.4°, 74.6°].
b.
SOLUTION: On the interval [0, 2π), the equation sin x = 1 has a
solution of
and the equation sin x =
solutions of and has .
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.
If the distance to the net is 50 feet, then the angle
would be 19.9° or 70.1°.
20. SKIING In the Olympics aerial skiing competition,
skiers speed down a slope that launches them into
the air, as shown. The maximum height a skier can
reach is given by h peak =
, where g is
the acceleration due to gravity or 9.8 meters per
second squared.
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.
a. If a skier obtains a height of 5 meters above the
end of the ramp, what was the skier’s initial speed?
b. Use your answer from part a to determine how
long it took the skier to reach the maximum height if
tpeak =
.
SOLUTION: a.
The interval is [15.4°, 74.6°].
b.
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Page 5
The skier’s initial speed was 11.67 meters per
second.
5-3
If
the distance
to the net is 50 feet,
then the angle
Solving
Trigonometric
Equations
would be 19.9° or 70.1°.
20. SKIING In the Olympics aerial skiing competition,
skiers speed down a slope that launches them into
the air, as shown. The maximum height a skier can
reach is given by h peak =
, where g is
The time it took the skier to reach the maximum
height was about 1.0 seconds.
Find all solutions of each equation on the
interval [0, 2 ).
21. 1 = cot 2 x + csc x
SOLUTION: the acceleration due to gravity or 9.8 meters per
second squared.
a. If a skier obtains a height of 5 meters above the
end of the ramp, what was the skier’s initial speed?
b. Use your answer from part a to determine how
long it took the skier to reach the maximum height if
tpeak =
.
SOLUTION: a.
Therefore, on the interval [0, 2π) the solutions are
,
The skier’s initial speed was 11.67 meters per
second.
b.
The time it took the skier to reach the maximum
height was about 1.0 seconds.
, and
.
22. sec x = tan x + 1
SOLUTION: Find all solutions of each equation on the
interval [0, 2 ).
21. 1 = cot 2 x + csc x
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SOLUTION: Page 6
Therefore, on the interval [0, 2π) the solutions are
5-3
,
, and
.
Solving
Trigonometric
Equations
is an extraneous solution. Therefore, on the
interval [0, 2π) the only solution is 0.
23. tan 2 x = 1 – sec x
22. sec x = tan x + 1
SOLUTION: SOLUTION: is an extraneous solution. Therefore, on the
interval [0, 2π) the only solution is 0.
Therefore, on the interval [0, 2π) the solutions are 0,
23. tan 2 x = 1 – sec x
, and
.
SOLUTION: 24. csc x + cot x = 1
SOLUTION: Therefore,
the interval
[0,
eSolutions
Manual - on
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by Cognero
, and
.
2π) the solutions are 0,
Page 7
Therefore, on the interval [0, 2π) the solutions are 0,
5-3 Solving
Equations
, and Trigonometric
.
x=
is an extraneous solution. Therefore, on the
interval [0, 2π) the only solution is
.
25. 2 – 2 cos2 x = sin x + 1
24. csc x + cot x = 1
SOLUTION: SOLUTION: Check
x=
is an extraneous solution. Therefore, on the
interval [0, 2π) the only solution is
.
25. 2 – 2 cos2 x = sin x + 1
SOLUTION: Therefore, on the interval [0, 2π) the solutions are
,
, and
.
26. cos x – 4 = sin x – 4
SOLUTION: Check
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Page 8
Therefore, on the interval [0, 2π) the solutions are
5-3 Solving
Trigonometric
Equations
,
, and
.
Therefore, on the interval [0, 2π) the only valid
solutions are
and
26. cos x – 4 = sin x – 4
27. 3 sin x = 3 – 3 cos x
SOLUTION: SOLUTION: .
Therefore, on the interval [0, 2π) the only valid
solutions are
and 0.
28. cot2 x csc2 x – cot2 x = 9
SOLUTION: Therefore, on the interval [0, 2π) the only valid
solutions are
and
.
27. 3 sin x = 3 – 3 cos x
is undefined. The solutions of
SOLUTION: are Check
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Page 9
are On the interval
, there are no real values of x
for which
, but
for 5-3 Solving Trigonometric Equations
Check
29. sec2 x – 1 + tan x –
tan x =
SOLUTION: Therefore, on the interval [0, 2π) the solutions are
,
,
and
.
30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3
SOLUTION: On the interval
, there are no real values of x
for which
, but
29. sec2 x – 1 + tan x –
for tan x =
SOLUTION: eSolutions Manual - Powered by Cognero
Page 10
interval [0, 2π) at
Therefore, on the interval [0, 2π) the solutions are
and .
Therefore, the components will be equivalent when
5-3 Solving
Equations
,
, Trigonometric
and
.
.
30. sec2 x tan2 x + 3 sec2 x – 2 tan2 x = 3
Find all solutions of each equation on the
interval [0, 2 ).
SOLUTION: 32. + cos x = 2
SOLUTION: The square of any real number must be greater than
2
or equal to zero so sec x = –1 has no solutions.
Therefore, on the interval [0, 2π) the only solutions
are 0 and π.
On the interval [0, 2π), cos x =
31. OPTOMETRY Optometrists sometimes join two
when x =
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 33. and .
= –4
+
and PV.
when x =
SOLUTION: 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
and .
On the interval [0, 2π),cos x = –
Therefore, the components will be equivalent when
.
Find all solutions of each equation on the
interval [0, 2 ).
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32. + cos x = 2
SOLUTION: and when x =
34. +
SOLUTION: when x =
.
= cos x
Page 11
On the interval [0, 2π),cos x = –
when x =
On
5-3 Solving
Trigonometric
Equations
and when x
=
.
34. +
when x =
,
and when x =
.
GRAPHING CALCULATOR Solve each
equation on the interval [0, 2 ) by graphing.
Round to the nearest hundredth.
= cos x
SOLUTION: 36. 3 cos 2x = ex + 1
SOLUTION: On the interval
0.33.
On the interval [0, 2π) there are no values of x for
which sin x = –2 and the sin x = 1 only when x =
However, tan
37. sin
.
x + cos
, the only solution is when x =
x = 3x
SOLUTION: is undefined, so there are no solutions for the original equation.
35. cot x cos x + 1 =
+ SOLUTION: On the interval
0.41.
, the only solution is when x =
38. x2 = 2 cos x + x
SOLUTION: On the interval
1.34.
On
,
when x =
and when x =
.
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Manual - Powered
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GRAPHING
CALCULATOR
Solve each
equation on the interval [0, 2 ) by graphing.
Round to the nearest hundredth.
, the only solution is when x =
39. x log x + 5x cos x = –2
SOLUTION: Page 12
cos
the interval
, the only Equations
solution is when x =
5-3 On
Solving
Trigonometric
1.34.
+ 66.95. Because x = 1 represents
January 15, x = 1.5 represents January 31. Use the
TRACE feature to determine the temperature when
x = 1.5.
39. x log x + 5x cos x = –2
SOLUTION: Therefore, the temperature on January 31 is about
61.3º.
b. Graph y = 70. Then use the intersect feature
under the CALC menu to determine the interval on
which 8.05 cos
+ 66.95 > 70.
On the interval
, the solutions are when x =
1.84 and when x = 4.49.
40. METEOROLOGY The average daily temperature
in degrees Fahrenheit for a city can be modeled by t
= 8.05 cos
+ 66.95, where x is a function of
time, x = 1 represents January 15, x = 2 represents
February 15, and so on.
a. Use a graphing calculator to estimate the
temperature on January 31.
b. Approximate the number of months that the
average daily temperature is greater than 70
throughout the entire month.
c. Estimate the highest temperature of the year and
the month in which it occurs.
SOLUTION: a. Use a graphing calculator to graph t = 8.05
cos
+ 66.95. Because x = 1 represents
January 15, x = 1.5 represents January 31. Use the
TRACE feature to determine the temperature when
x = 1.5.
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Therefore, the temperature on January 31 is about
61.3º.
So, 8.05 cos
+ 66.95 > 70 on the interval
3.74 < x < 8.26. Because x = 3.74 corresponds to
about the middle of April, the temperature does not
exceed 70º for the entire month until May.
Therefore, the average daily temperature is greater
than 70º for the entire month during the months of
May, June, and July. Therefore, the average
temperature was higher than 70º for three months.
c. Use the maximum feature under the CALC
menu to find the maximum value of the function on
[0, 12].
Page 13
The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º
May, June, and July. Therefore, the average
temperature was higher than 70º for three months.
On the interval [0, 2π) cos x = 0 when x =
and x
Use the maximum
feature under
the CALC
5-3 c.
Solving
Trigonometric
Equations
=
.
menu to find the maximum value of the function on
[0, 12].
42. The maximum value is 75 when x ≈ 6. Therefore, the maximum average monthly temperature is 75º
during the month of June.
SOLUTION: Let y = 0 and solve for x.
Find the x-intercepts of each graph on the
interval [0, 2 ).
On the interval [0, 2π) sin x = 0 when x = 0 and x =
π.
41. SOLUTION: Let y = 0 and solve for x.
43. On the interval [0, 2π) cos x = 0 when x =
and x
SOLUTION: Let y = 0 and solve for x.
=
.
42. On the interval [0, 2π) cot x = 1 when x =
=
and x
.
SOLUTION: Let y = 0 and solve for x.
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Page 14
44. and x
On the interval [0, 2π) cot x = 1 when x =
On the interval [0, 4π) the solutions are
, and
5-3 =Solving
Trigonometric Equations
.
,
,
.
46. 2 sin2 x + 1 = –3 sin x
SOLUTION: 44. SOLUTION: Let y = 0 and solve for x.
On the interval [0, 4π) the solutions are
,
,
, and
,
,
.
47. csc x cot2 x = csc x
On the interval [0, 2π) tan x = 0 when x = 0 and x =
π, tan x = –1 when x =
and x =
and x =
= 1 when x =
, and tan x
SOLUTION: .
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
On the interval [0, 4π) the solutions are
, and
,
,
,
.
,
,
,
, and
,
,
,
,
.
48. sec x + 5 = 2 sec x + 3
SOLUTION: 46. 2 sin2 x + 1 = –3 sin x
SOLUTION: On the interval [0, 4π) the solutions are
On the interval [0, 4π) the solutions are
,
,
, and
.
eSolutions Manual - Powered by Cognero
,
,
, and
.
Page 15
49. GEOMETRY Consider the circle below.
On the interval [0, 4π) the solutions are
,
,
5-3 Solving
,
, Trigonometric
,
,
, and Equations
.
On the graphing calculator, find the intersection of
Y1 = 18sinθ and Y2 = 14θ.
48. sec x + 5 = 2 sec x + 3
SOLUTION: On the interval [0, 4π) the solutions are
, and
,
,
The value of 2θ = 2(1.1968) or about 2.39 radians.
b. First, substitute into the given area formula and
rearrange it.
.
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
=
Using a graphing calculator, find the intersection of
Y1 = 2.88 and Y2 = θ – sin θ.
. 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
=
.
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 ).
50. 1 > 2 sin x
SOLUTION: Graph y = 1 and y = 2 sin x on [0, 2π). Use the
intersect feature under the CALC menu to
determine on what interval(s) 1 > 2 sin x.
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 16
graph when x <
and when x >
> 2 sin x on 0 ≤ x < or the area
is 36 square inches
and the radius is
5-3 When
Solving
Trigonometric
Equations
. Therefore, 1
< x < 2π.
5 inches, then the measure of θ is 3.01 radians.
Solve each inequality on the interval [0, 2 ).
50. 1 > 2 sin x
51. 0 < 2 cos x –
SOLUTION: SOLUTION: Graph y = 2 cos x –
on [0, 2π). Use the zero
CALC
feature under the
menu to determine on
what interval(s) 0 < 2 cos x –
.
Graph y = 1 and y = 2 sin x on [0, 2π). Use the
intersect feature under the CALC menu to
determine on what interval(s) 1 > 2 sin x.
The graphs intersect at about 0.524 or
2.618 or
The zeros of y = 2 cos x –
and about . The graph of y = 1 is above the other
graph when x <
and when x >
> 2 sin x on 0 ≤ x < or and about 5.498 or . Therefore, 1
. The graph of y = 0 is
below the other graph when 0 ≤ x <
Therefore, 0 < 2 cos x –
< x <
on 0 ≤ x <
or < x < 2π.
51. 0 < 2 cos x –
Graph y = 2 cos x –
on [0, 2π). Use the zero
feature under the CALC menu to determine on
what interval(s) 0 < 2 cos x –
.
or 2π.
< x < 2π.
SOLUTION: are about 0.785 or
52. ≥ SOLUTION: Graph y =
and y =
on [0, 2π). Use
the intersect feature under the CALC menu to
determine on what interval(s)
eSolutions Manual - Powered by Cognero
≥ .
Page 17
below the other graph when 0 ≤ x <
or other graph when < x <
≤ x ≤ 2π.
Therefore, 0 < 2 cos x –
on 0 ≤ x <
or < 5-3 Solving Trigonometric Equations
53. ≥ on
≤ x ≤ .
≤ tan x cot x
SOLUTION: SOLUTION: Graph y =
≥ Therefore,
x < 2π.
52. .
and y =
on [0, 2π). Use
the intersect feature under the CALC menu to
≥ determine on what interval(s)
.
Graph y =
and y = tan x cot x on [0, 2π).
Use the intersect feature under the CALC menu to
≤ tan x cot
determine on what interval(s)
x.
The graphs intersect at about 3.142 or π. The graph
of y = tan x cot x is at or below the other graph
when 0 ≤ x < 2π.
The graphs intersect at about 1.047 or
2.094 or
. The graph of y =
Therefore,
and about is below the
.
Graph y = cos x and y =
≥ Therefore,
54. cos x ≤
SOLUTION: other graph when ≤ x ≤ ≤ tan x cot x on 0 ≤ x < 2π.
on
≤ x ≤ .
on [0, 2π). Use the
intersect feature under the CALC menu to
determine on what interval(s) cos x ≤
53. .
≤ 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.
eSolutions Manual - Powered by Cognero
Page 18
of y = tan x cot x is at or below the other graph
when 0 ≤ x < 2π.
other graph when
5-3 Therefore,
Solving Trigonometric
Equations
x on 0 ≤ x < 2π.
≤ tan x cot
Therefore, cos x ≤
55. 54. cos x ≤
≤ x ≤ on
.
≤ x ≤ .
sin x − 1 < 0
SOLUTION: SOLUTION: on [0, 2π). Use the
Graph y = cos x and y =
Graph y =
sin x − 1. Use the zero feature under
CALC
the
menu to determine on what interval(s)
sin x − 1 < 0.
intersect feature under the CALC menu to
determine on what interval(s) cos x ≤
.
The zeros of
about 2.356 or
when 0 ≤ x <
The graphs intersect at about 2.618 or
3.665 or
and about
. The graph of y = cos x is below the
other graph when
≤ x ≤ and
. The graph is below the x-axis
or < x < 2π.
sin x − 1 < 0 on 0 ≤ x <
or < x < 2π.
56. REFRACTION When light travels from one
.
medium to another it bends or refracts, as shown.
Therefore, cos x ≤
55. Therefore,
sin x − 1 are about 0.785 or
on
≤ x ≤ .
sin x − 1 < 0
SOLUTION: Graph y =
sin x − 1. Use the zero feature under
the CALC menu to determine on what interval(s)
sin x − 1 < 0.
Refraction is described by n 2 sinθ1 = n 1 sinθ2, where
n 1 is the index of refraction of the medium the light
is entering, n 2 is the index of refraction of the
medium the light is exiting, θ1 is the angle of
incidence, and θ2 is the angle of refraction.
a. Find θ2 for each material shown if the angle of
incidence is 40º and the index of refraction for air is
1.00.
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Page 19
medium the light is exiting, θ1 is the angle of
incidence, and θ2 is the angle of refraction.
a. Find θ for each material shown if the angle of
2
5-3 Solving
Trigonometric Equations
incidence is 40º and the index of refraction for air is
1.00.
water:
b. If the angle of incidence is doubled to 80 , will
the resulting angles of refraction be twice as large as
those found in part a?
b. Find the angle of refraction for one of the
mediums using an angle of incidence of 80 .
glass:
SOLUTION: a. glass:
ice:
θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 =
80 . Therefore, the angle of refraction is not twice
as large.
57. ERROR ANALYSIS Vijay and Alicia are solving
2
tan x – tan x +
= the solutions are x =
+ n , and x =
plastic:
solutions are x =
tan x. Vijay thinks that
+ n , x =
+ n , x =
+ n . Alicia thinks that the
+ n
+ n . Is
and x =
either of them correct? Explain your reasoning.
SOLUTION: 2
First, solve tan x – tan x +
water:
= tan x.
On [0, 2π) tan x = 1 when x =
tan x =
when x =
and x =
and x =
and .
Sample answer: Therefore, Vijay’s solutions are
correct; however, they are not stated in the simplest
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b. Find the angle of refraction for one of the
mediums using an angle of incidence of 80 .
Page 20
form. For example, his solutions of x =
+ n
and
θ2 ≈ 25 when θ1 = 40 andθ2 ≈ 40.4 when θ1 =
n
. Therefore,
the angle of refraction
is not twice
5-3 80
Solving
Trigonometric
Equations
.
as large.
57. ERROR ANALYSIS Vijay and Alicia are solving
2
tan x – tan x +
= tan x. Vijay thinks that
+ n , x =
the solutions are x =
+ n , x =
+ nπ is equivalent to
because when n = 1,
CHALLENGE Solve each equation on the
interval [0, 2π].
58. 16 sin5 x + 2 sin x = 12 sin3 x
SOLUTION: + n , and x =
solutions are x =
+ n . Alicia thinks that the
+ n
+ n . Is
and x =
either of them correct? Explain your reasoning.
SOLUTION: 2
First, solve tan x – tan x +
= tan x.
On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =
x=
when x =
and x =
, sin x =
and x =
, sin x = –
when x =
when
and On [0, 2π) tan x = 1 when x =
tan x =
when x =
and x =
and x =
and x=
when x =
, and sin x =
and x =
.
Therefore, after checking for extraneous solutions,
.
Sample answer: Therefore, Vijay’s solutions are
correct; however, they are not stated in the simplest
form. For example, his solutions of x =
+ n
x=
n
+ n
could simply be stated as x =
because when n = 1,
the solutions are 0,
, and
and
+ + nπ is equivalent to
,
,
,
,
,
,
,
.
59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3
SOLUTION: .
CHALLENGE Solve each equation on the
interval [0, 2π].
58. 16 sin5 x + 2 sin x = 12 sin3 x
SOLUTION: On [0, 2 ) sin x = 1 when x =
x=
, sin x = –
and sin x =
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On [0, 2π) sin x = 0 when x = 0 and x = π, sin x =
when x =
and x =
, sin x =
when
, sin x = –1 when
and x =
when x =
when x =
and x =
,
.
Page 21
Therefore, after checking for extraneous solutions,
the solutions are
,
,
,
,
, and
.
the solutions are 0,
,
,
,
,
,
,
,
Therefore, after checking for extraneous solutions,
, and Trigonometric
.
5-3 Solving
Equations
the solutions are
59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3
,
,
,
,
, and
.
60. REASONING Are the solutions of csc x =
and
2
cot x + 1 = 2 equivalent? If so, verify your answer
algebraically. If not, explain your reasoning.
SOLUTION: SOLUTION: On [0, 2 ) sin x = 1 when x =
x=
, sin x = –
and sin x =
, sin x = –1 when
and x =
when x =
when x =
and x =
,
.
Therefore, after checking for extraneous solutions,
the solutions are
,
,
,
,
, and
60. REASONING Are the solutions of csc x =
.
are The solutions to csc x =
and
and . Using
2
2
cot x + 1 = 2 equivalent? If so, verify your answer
algebraically. If not, explain your reasoning.
a Pythagorean identity, cot x + 1 = 2 simplifies to
SOLUTION: two additional solutions when csc x = –
2
csc x = 2 or csc x = ±
. Therefore, there are
. So, the solutions of csc x =
and
:
2
and cot x + 1
= 2 are not equivalent.
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.
eSolutions
- Powered
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solutions
to cscbyxCognero
=
are and . Using
When sin x =
,x=
and x =
Page 22
. So, one
2
a Pythagorean identity, cot x + 1 = 2 simplifies to
2
equation that has solutions of 0, π,
, and
is two additional solutions when csc x = –
and
:
2
is and
= 0 or cos x
2
and cot x + 1
. So, the solutions of csc x =
5-3 Solving Trigonometric Equations
–
= 2 are not equivalent.
OPEN ENDED Write a trigonometric equation
that has each of the following solutions.
2
= 0. This can be rewritten as 4 cos x = 3.
63. Writing in Math Explain the difference in the
techniques that are used when solving equations and
verifying identities.
SOLUTION: When solving an equation, you use properties of
equality to manipulate each side of the equation to
isolate a variable. For example:
61. SOLUTION: Sample answer: When sin x = 0, x = 0 and x = π.
When sin x =
and x =
,x=
. So, one
equation that has solutions of 0, π,
2
2
can be rewritten as 2 sin x =
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.For example:
is , and
= 0 or sin x –
= 0. This
Verify
sin x.
=
.
62. SOLUTION: Sample answer: When cos x =
. When cos x = –
,x=
and x =
,x=
and x =
So, one equation that has solutions of
,
Verify each identity.
.
,
,
64. =
SOLUTION: and
–
is 2
= 0 or cos x
2
= 0. This can be rewritten as 4 cos x = 3.
63. Writing in Math Explain the difference in the
techniques that are used when solving equations and
verifying
eSolutions
Manualidentities.
- Powered by Cognero
SOLUTION: When solving an equation, you use properties of
Page 23
+
5-3 Solving Trigonometric Equations
Verify each identity.
64. 2
2
= cos θ + sin θ Reciprocal Identities
= 1 Pythagorean Identity
Find the value of each expression using the
given information.
=
67. tan
; sin θ =
, tan
> 0
SOLUTION: SOLUTION: Use the Pythagorean Identity that involves sin θ.
Since tan
is positive and sin is positive, cos θ
must be positive. So,
65. .
=
SOLUTION: 66. = 1
+
SOLUTION: +
2
2
= cos θ + sin θ Reciprocal Identities
= 1 Pythagorean Identity
68. csc
, cos = , csc
< 0
SOLUTION: Find the value of each expression using the
given information.
67. tan
; sin θ =
, tan
Use the Pythagorean Identity that involves cos θ.
> 0
SOLUTION: Use the Pythagorean Identity that involves sin θ.
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Page 24
in years.
a. What is the amplitude of the function? What does
it represent?
b. What is the period of the function? What does it
represent?
c. Graph the function.
5-3 Solving Trigonometric Equations
68. csc
, cos = , csc
< 0
SOLUTION: a. The amplitude of the function is |a| = |20,000| or
20,000. The amplitude represents the amount that the
population varies above and below the initial
population of 30,000.
SOLUTION: Use the Pythagorean Identity that involves cos θ.
b. The period of the function is
The population will return to its original value every
20 years.
Create a table listing the coordinates of the xintercepts and extrema for f (x) = sin x for one
period, 2 , on the interval [0, 2 ]. Then use the
amplitude of g(x) to find corresponding points on its
graph.
Functions
Max
Min
Max
Since one of the conditions is that csc θ < 0,
.
69. sec
; tan
= –1, sin
< 0
SOLUTION: Use the Pythagorean Identity that involves tan
(0, 50)
.
(10, 10)
(20, 50)
Sketch the curve through the indicated points for the
function. Then repeat the pattern to complete a
second period.
Since tan is negative and sin θ is negative, cos
and sec θ must be positive. Therefore,
.
70. POPULATION The population of a certain species
of deer can be modeled by p = 30,000 + 20,000
cos
, where p is the population and t is the time
in years.
a. What is the amplitude of the function? What does
it represent?
b. What is the period of the function? What does it
represent?
c. Graph the function.
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SOLUTION: a. The amplitude of the function is |a| = |20,000| or
Given f (x) = 2x 2 – 5x + 3 and g(x) = 6x + 4, find
each of the following.
71. (f – g)(x)
SOLUTION: Page 25
5-3 Solving Trigonometric Equations
Given f (x) = 2x 2 – 5x + 3 and g(x) = 6x + 4, find
each of the following.
71. (f – g)(x)
74. BUSINESS A small business owner must hire
seasonal workers as the need arises. The following
list shows the number of employees hired monthly
for a 5-month period.
SOLUTION: If the mean of these data is 9, what is the population
standard deviation for these data? Round to the
nearest tenth.
72. (f ⋅ g)(x)
SOLUTION: 73. SOLUTION: To find the standard deviation, first find the variance.
The mean of the data is 9. Use the mean to find the
variance.
(x)
SOLUTION: Now find the standard deviation by taking the square
root of the variance.
The standard deviation is about 3.5.
75. SAT/ACT For all positive values of m and n, if
= 2, then x =
A
74. BUSINESS A small business owner must hire
seasonal workers as the need arises. The following
list shows the number of employees hired monthly
for a 5-month period.
B
C
D
If the mean of these data is 9, what is the population
standard deviation for these data? Round to the
nearest tenth.
E
SOLUTION: SOLUTION: To find the standard deviation, first find the variance.
The mean of the data is 9. Use the mean to find the
variance.
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Page 26
The correct answer is D.
Now find the standard deviation by taking the square
root of the variance.
5-3 Solving Trigonometric Equations
The standard deviation is about 3.5.
The correct answer is H.
75. SAT/ACT For all positive values of m and n, if
77. Which of the following is not a solution of 0 = sin
2
+ cos tan
= 2, then x =
?
A
A
B
B
C 2π
C
D
D
SOLUTION: Try choice A.
E
SOLUTION: Try choice B.
The correct answer is D.
76. If cos x = –0.45, what is sin
?
Try choice C.
F – 0.55
G – 0.45
H 0.45
J 0.55
Try choice D.
SOLUTION: The correct answer is D.
78. REVIEW The graph of y = 2 cos
The correct answer is H.
Which is a solution for 2 cos
is shown. = 1?
77. Which of the following is not a solution of 0 = sin
2
+ cos tan
?
A
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eSolutions
C 2π
Page 27
F
5-3 Solving Trigonometric Equations
The correct answer is D.
78. REVIEW The graph of y = 2 cos
Which is a solution for 2 cos
is shown. = 1?
F
G
H
J
SOLUTION: Choice F:
No
Choice G:
No
Choice H:
Choice J:
Yes
No
The correct answer is H.
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