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Lesson 7.2 Exercises, pages 592–599
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 < 2␲.
√
3
1
a) sin x = 2
b) tan x = √
3
The reference angle is:
The reference angle is:
√
ⴚ1
sin
3
␲
a b ⴝ
2
3
In Quadrant 1, x ⴝ
tan–1 a√ b ⴝ
1
3
␲
3
In Quadrant 2, x ⴝ ␲ ⴚ
or
6
2␲
3
␲
6
In Quadrant 1, x ⴝ
␲
,
3
␲
6
In Quadrant 3, x ⴝ ␲ ⴙ
7.2 Solving Trigonometric Equations Algebraically—Solutions
7␲
␲
, or
6
6
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5. Verify that each given value of x is a root of the equation.
a) tan2x ⫺ 3 ⫽ 0;
␲
b) 8 sin2x ⫹ 6 sin x ⫹ 1 ⫽ 0;
7␲
x = 3
x = 6
␲
Substitute x ⴝ in each side
3
of the equation.
␲
L.S. ⴝ tan2 a b ⴚ 3
3
√ 2
ⴝ ( 3) ⴚ 3
ⴝ0
ⴝ R.S.
Since the left side is equal to
the right side, the root is
verified.
7␲
Substitute x ⴝ
in each side of
6
the equation.
L.S. ⴝ 8 sin2 a
7␲
7␲
ⴙ1
b ⴙ 6 sin
6
6
2
ⴝ 8 aⴚ b ⴙ 6 aⴚ b ⴙ 1
1
2
1
2
ⴝ2ⴚ3ⴙ1
ⴝ0
ⴝ R.S.
Since the left side is equal to the
right side, the root is verified.
B
6. Solve each equation over the domain 0 ≤ x < 2␲, then state the
general solution.
a) 3 cos x ⫺ 2 ⫽ 0
3 cos x ⴝ 2
cos x ⴝ
2
3
In Quadrant 1,
x ⴝ cosⴚ1 a b
2
3
x ⴝ 0.8410. . .
In Quadrant 4,
x ⴝ 2␲ ⴚ 0.8410. . .
x ⴝ 5.4421. . .
The roots are: x ⬟ 0.84 and
x ⬟ 5.44
The general solution is:
x ⬟ 0.84 ⴙ 2␲k, k ç ⺪ or
x ⬟ 5.44 ⴙ 2␲k, k ç ⺪
©P DO NOT COPY.
b) 2 tan x ⫹
√
5⫽0
√
2 tan x ⴝ ⴚ 5
√
tan x ⴝ ⴚ
5
2
The reference angle is:
√
tan a
ⴚ1
5
b ⴝ 0.8410. . .
2
In Quadrant 2,
x ⴝ ␲ ⴚ 0.8410. . .
x ⴝ 2.3005. . .
In Quadrant 4,
x ⴝ 2␲ ⴚ 0.8410. . .
x ⴝ 5.4421. . .
The roots are: x ⬟ 2.30 and
x ⬟ 5.44
The general solution is:
x ⬟ 2.30 ⴙ ␲k, k ç ⺪ or
x ⬟ 5.44 ⴙ ␲k, k ç ⺪
7.2 Solving Trigonometric Equations Algebraically—Solutions
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7. Solve each equation for -␲ ≤ x ≤ ␲.
a) 3 tan x ⫺ 3 ⫽ 5 tan x ⫺ 1
b) 5(1 ⫹ 2 sin x) ⫽ 2 sin x ⫹ 1
2 tan x ⴝ ⴚ2
tan x ⴝ ⴚ1
The reference angle is:
5 ⴙ 10 sin x ⴝ 2 sin x ⴙ 1
8 sin x ⴝ ⴚ4
sin x ⴝ ⴚ
1
2
␲
tan (1) ⴝ
4
The reference angle is:
In Quadrant 2,
␲
xⴝ␲ⴚ
sinⴚ1 a b ⴝ
ⴚ1
xⴝ
3␲
4
1
2
4
In Quadrant 3,
In Quadrant 4,
x ⴝ ⴚ␲ ⴙ
xⴝⴚ
xⴝⴚ
␲
4
The roots are: x ⴝ
␲
xⴝⴚ
4
␲
6
3␲
and
4
5␲
6
␲
6
In Quadrant 4,
xⴝⴚ
␲
6
␲
6
The roots are: x ⴝ ⴚ and
xⴝⴚ
5␲
6
8. a) Solve each equation for ⫺180° ≤ x ≤ 90°.
i) 2 csc x ⫽ 6
ii) ⫺6 ⫽ 3 cot x
csc x ⴝ 3
ⴚ2 ⴝ cot x
1
sin x ⴝ
3
tan x ⴝ ⴚ
The reference angle is:
The reference angle is:
1
sinⴚ1 a b ⴝ 19.4712. . . º
3
tan–1 a b ⴝ 26.5650. . . º
In Quadrant 1, x ⴝ 19.4712. . . º
Quadrant 2 is not in the domain.
The root is: x ⬟ 19°
Quadrant 2 is not in the domain.
In Quadrant 4, x ⴝ ⴚ 26.5650. . . º
The root is: x ⬟ ⴚ27º
1
2
1
2
b) Solve each equation for ⫺90° ≤ x ≤ 180°.
i) 4 sec x ⫽ ⫺5
ⴚ5
4
4
cos x ⴝ ⴚ
5
8
1
1
ii) ⫺2 ⫽ 3 csc x
3
2
2
sin x ⴝ ⴚ
3
sec x ⴝ
csc x ⴝ ⴚ
The reference angle is:
The reference angle is:
4
cosⴚ1 a b ⴝ 36.8698. . . °
5
sinⴚ1 a b ⴝ 41.8103. . . °
In Quadrant 2,
x ⴝ 180° ⴚ 36.8698. . . °
x ⴝ 143.1301. . . °
There is no solution in Quadrant 3.
The root is: x ⬟ 143°
There is no solution in Quadrant 3.
In Quadrant 4,
x ⴝ ⴚ41.8103. . . °
The root is: x ⬟ ⴚ42°
2
3
7.2 Solving Trigonometric Equations Algebraically—Solutions
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9. For each equation, determine the general solution over the set of
real numbers, then list the roots over the domain ⫺␲ ≤ x < 0.
a) cos 3x ⫺ 1 ⫽ 5 cos 3x ⫹ 2
b) 3 sin 4x ⫽ 3 ⫺ 2 sin 4x
5 sin 4x ⴝ 3
ⴚ3 ⴝ 4 cos 3x
3
cos 3x ⴝ ⴚ
4
sin 4x ⴝ
3x ⴝ cosⴚ1 aⴚ b
3
4
3
5
4x ⴝ sinⴚ1 a b
3
5
The terminal arm of angle 3x lies
in Quadrant 2 or 3.
The reference angle for angle 3x is:
The terminal arm of angle 4x lies in
Quadrant 1 or 2.
The reference angle for angle 4x is:
cosⴚ1 a b ⴝ 0.7227. . .
sinⴚ1 a b ⴝ 0.6435 . . .
In Quadrant 2,
3x ⴝ ␲ ⴚ 0.7227. . .
x ⴝ 0.8062. . .
In Quadrant 3,
3x ⴝ ␲ ⴙ 0.7227. . .
x ⴝ 1.2881. . .
2␲
The period of cos 3x is , so the
3
general solution is:
In Quadrant 1,
4x ⴝ 0.6435. . .
x ⴝ 0.1608. . .
In Quadrant 2,
4x ⴝ ␲ ⴚ 0.6435. . .
x ⴝ 0.6245. . .
x ⬟ 0.81 ⴙ
x ⬟ 0.16 ⴙ
In the given domain, the roots are:
In the given domain, the roots are:
2␲
x ⴝ 0.8062. . . ⴚ
3
x ⴝ 0.1608. . . ⴚ
x ⬟ ⴚ1.29, and
x ⬟ ⴚ1.41, and
x ⴝ 0.1608. . . ⴚ ␲
x ⬟ ⴚ2.98, and
3
4
2␲
k, k ç ⺪, or
3
2␲
x ⬟ 1.29 ⴙ
k, k ç ⺪
3
2␲
x ⴝ 1.2881. . . ⴚ
3
x ⬟ ⴚ0.81, and
x ⴝ 1.2881. . . ⴚ
x ⬟ ⴚ2.90
4␲
3
3
5
The period of sin 4x is
␲
2␲
, or ,
4
2
so the general solution is:
␲
k, k ç ⺪, or
2
␲
x ⬟ 0.62 ⴙ k, k ç ⺪
2
x ⴝ 0.6245. . . ⴚ
␲
2
␲
2
x ⬟ ⴚ0.95, and
x ⴝ 0.6245. . . ⴚ ␲
x ⬟ ⴚ2.52
10. Two students determined the general solution of the equation
3 sin x ⫹ 5 ⫽ 5(sin x ⫹ 1). Joseph said the solution is x ⫽ 2␲k or
x ⫽ ␲ ⫹ 2␲k, where k is an integer. Yeoun Sun said the solution
is x ⫽ ␲k, where k is an integer. Who is correct? Explain.
Both students are correct because both expressions produce the same
roots: x ⴝ 0, x ⴝ — ␲, x = — 2␲, x ⴝ — 3␲, and so on
©P DO NOT COPY.
7.2 Solving Trigonometric Equations Algebraically—Solutions
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␲
11. Solve each equation over the domain ⫺␲ ≤ x ≤ 2 .
a) 4 cos2x ⫺ 3 ⫽ 0
b) 2 tan2x ⫽ 3
4 cos2x ⴝ 3
cos2x ⴝ
3
4
cos x ⴝ —
tan2x ⴝ
tan x ⴝ —
√
3
2
√3
√3
2
The reference angle is:
tan a
The reference angle is:
ⴚ1
√
cosⴚ1 a
3
2
3
␲
bⴝ
2
6
2
b ⴝ 0.8860. . .
In Quadrant 1, x ⴝ 0.8860. . .
In Quadrant 2, there is no solution in
the given domain.
In Quadrant 3, x ⴝ ⴚ␲ ⴙ 0.8860. . .
x ⴝ ⴚ2.2555. . .
In Quadrant 4, x ⴝ ⴚ0.8860. . .
The roots are: x ⬟ — 0.89 and
x ⬟ ⴚ2.26
␲
In Quadrant 1, x ⴝ
6
In Quadrant 2, there is no
solution in the given domain.
␲
In Quadrant 3, x ⴝ ⴚ␲ ⴙ
6
5␲
6
␲
In Quadrant 4, x ⴝ ⴚ
6
␲
The roots are: x ⴝ — and
6
5␲
xⴝⴚ
6
xⴝⴚ
12. Use factoring to solve each equation over the domain
⫺90° ≤ x < 270°.
a) 2 cos x sin x ⫺ cos x ⫽ 0
b) 3 tan x ⫹ tan2x ⫽ 2 tan x
(cos x)(2 sin x ⴚ 1) ⴝ 0
Either cos x ⴝ 0
x ⴝ — 90°
Or 2 sin x ⴚ 1 ⴝ 0
sin x ⴝ 0.5
x ⴝ 30º
or x ⴝ 180º ⴚ 30º
x ⴝ 150º
The roots are: x ⴝ 30º,
x ⴝ — 90°, and x ⴝ 150º
tan x ⴙ tan2x ⴝ 0
(tan x)(1 ⴙ tan x) ⴝ 0
Either tan x ⴝ 0
x ⴝ 0° or x ⴝ 180°
Or 1 ⴙ tan x ⴝ 0
tan x ⴝ ⴚ1
x ⴝ 135º or x ⴝ ⴚ45º
The roots are: x ⴝ 0º, x ⴝ 180°,
x ⴝ 135º, and x ⴝ ⴚ45º
13. A student wrote the solution below to solve the equation
2 sin2x ⫹ sin x ⫽ 1 over the domain 0 ≤ x < 2␲. Identify any
errors, then write a correct solution.
2 sin2x ⫹ sin x ⫽ 1
(sin x)(2 sin x ⫹ 1) ⫽ 1
sin x ⫽ 1
or
2 sin x ⫹ 1 ⫽ 1
x⫽
␲
2
2 sin2x ⴙ sin x ⴚ 1 ⴝ 0
(2 sin x ⴚ 1)(sin x ⴙ 1) ⴝ 0
Either 2 sin x ⴚ 1 ⴝ 0
sin x ⴝ
sin x ⫽ 0
x ⫽ 0 or x ⫽ ␲
xⴝ
␲
5␲
or x ⴝ
6
6
1
2
Or sin x ⴙ 1 ⴝ 0
sin x ⴝ ⴚ1
xⴝ
3␲
2
There is an error in the second line of the solution. One side of the equation
must be 0 before factoring, so collect all the terms on the left side.
10
7.2 Solving Trigonometric Equations Algebraically—Solutions
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14. Solve each equation over the domain ⫺2␲ ≤ x ≤ 2␲, then
determine the general solution.
a) 2 cos2x ⫺ cos x ⫺ 1 ⫽ 0
b) 5 sin2x ⫹ 3 sin x ⫽ 2
(2 cos x ⴙ 1)(cos x ⴚ 1) ⴝ 0
For 2 cos x ⴙ 1 ⴝ 0
2 cos x ⴝ ⴚ1
cos x ⴝ ⴚ
1
2
The terminal arm of angle x lies
in Quadrant 2 or 3.
The reference angle is:
cosⴚ1 a b ⴝ
1
2
␲
3
5 sin2x ⴙ 3 sin x ⴚ 2 ⴝ 0
(5 sin x ⴚ 2)(sin x ⴙ 1) ⴝ 0
For 5 sin x ⴚ 2 ⴝ 0
5 sin x ⴝ 2
sin x ⴝ
2
5
The terminal arm of angle x lies in
Quadrant 1 or 2.
The reference angle is:
sinⴚ1 a b ⴝ 0.4115. . .
2
5
In Quadrant 2,
2␲
␲
, or
3
3
4␲
␲
and, x ⴝ ⴚ␲ ⴚ , or ⴚ
3
3
For cos x ⴚ 1 ⴝ 0
cos x ⴝ 1
The terminal arm lies along the
positive x-axis.
So, x ⴝ — 2␲ or x ⴝ 0
In Quadrant 1,
x ⴝ 0.4115. . .
and, x ⴝ ⴚ2␲ ⴙ 0.4115. . .
x ⴝ ⴚ5.8716. . .
In Quadrant 2,
x ⴝ ␲ ⴚ 0.4115. . .
x ⴝ 2.7300. . .
and, x ⴝ ⴚ␲ ⴚ 0.4115. . .
x ⴝ ⴚ3.5531. . .
For sin x ⴙ 1 ⴝ 0
sin x ⴝ ⴚ1
The terminal arm lies along the
negative y-axis.
The roots are: x ⴝ 0, x ⴝ —
So, x ⴝ
xⴝ␲ⴚ
In Quadrant 3,
4␲
␲
, or
3
3
2␲
␲
and, x ⴝ ⴚ␲ ⴙ , or ⴚ
3
3
xⴝ␲ⴙ
xⴝ—
4␲
, and x ⴝ — 2␲
3
2␲
,
3
Since consecutive roots differ
2␲
, the general solution is:
3
2␲
x ⴝ k, k ç ⺪
3
by
The roots are: x ⬟ ⴚ5.87,
␲
2
3␲
x ⬟ 2.73, and x ⴝ
2
x ⬟ ⴚ3.55, x ⴝ ⴚ , x ⬟ 0.41,
Since the period is 2␲, the general
solution is:
x ⬟ 0.41 ⴙ 2␲k, k ç ⺪ or
x ⬟ 2.73 ⴙ 2␲k, k ç ⺪ or
xⴝ
©P DO NOT COPY.
␲
3␲
or x ⴝ ⴚ
2
2
3␲
ⴙ 2␲k, k ç ⺪
2
7.2 Solving Trigonometric Equations Algebraically—Solutions
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15. Solve each equation over the domain 0 ≤ x < 2␲.
a) 4 tan2x ⫽ 2 ⫺ 5 tan x
4 tan2x ⴙ 5 tan x ⴚ 2 ⴝ 0
√
Use: tan x ⴝ
ⴚ b_ b2 ⴚ 4ac
2a
b) 4 sin x ⫹ 3 ⫽ 2 sin2x
2 sin2x ⴚ 4 sin x ⴚ 3 ⴝ 0
Use: sin x ⴝ
√
ⴚ b_ b2 ⴚ 4ac
2a
Substitute: a ⴝ 4, b ⴝ 5,
c ⴝ ⴚ2
√
Substitute: a ⴝ 2, b ⴝ ⴚ4,
c ⴝ ⴚ3
√
tan x ⴝ
sin x ⴝ
ⴚ 5_ 52 ⴚ 4(4)( ⴚ 2)
4_ ( ⴚ 4)2 ⴚ 4(2)( ⴚ 3)
2(4)
√
ⴚ 5_ 57
tan x ⴝ
8
√
ⴚ 5 ⴙ 57
For tan x ⴝ
8
2(2)
√
√
2_ 10
4_ 40
sin x ⴝ
, or
4
2
√
2 ⴙ 10
For sin x ⴝ
; this is greater
2
tan x is positive when the
terminal arm of angle x lies in
Quadrant 1 or 3.
The reference angle is:
than 1, so there is no real solution.
ⴚ5 ⴙ
tanⴚ1 a
8
√
57
b ⴝ 0.3085. . .
In Quadrant 1, x ⴝ 0.3085. . .
In Quadrant 3,
x ⴝ ␲ ⴙ 0.3085. . .
x ⴝ 3.4501. . .
ⴚ5 ⴚ
For tan x ⴝ
8
√
57
tan x is negative when the
terminal arm of angle x lies in
Quadrant 2 or 4.
The reference angle is:
tanⴚ1 a
5ⴙ
√
57
b ⴝ 1.0032. . .
8
For sin x ⴝ
2ⴚ
√
10
2
sin x is negative when the
terminal arm of angle x lies
in Quadrant 3 or 4.
The reference angle is:
√
sinⴚ1 a
10 ⴚ 2
b ⴝ 0.6201. . .
2
In Quadrant 3,
x ⴝ ␲ ⴙ 0.6201. . .
x ⴝ 3.7617. . .
In Quadrant 4,
x ⴝ 2␲ ⴚ 0.6201. . .
x ⴝ 5.6630. . .
The roots are: x ⬟ 3.76 and
x ⬟ 5.66
In Quadrant 2,
x ⴝ ␲ ⴚ 1.0032. . .
x ⴝ 2.1383. . .
In Quadrant 4,
x ⴝ 2␲ ⴚ 1.0032. . .
x ⴝ 5.2798. . .
The roots are: x ⬟ 0.31,
x ⬟ 2.14, x ⬟ 3.45, and
x ⬟ 5.28
12
7.2 Solving Trigonometric Equations Algebraically—Solutions
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C
16. Write a second-degree trigonometric equation that has roots
␲ ␲ 5␲
6 , 2 , 6 over the domain 0 ≤ x < 2␲.
Sample response: Since sin
␲
5␲
ⴝ sin , these two roots can be determined
6
6
from one factor, so use the sine ratio. Work backward.
␲
6
1
So, sin x ⴝ
2
Either x ⴝ
Then, sin x ⴚ
1
ⴝ0
2
␲
2
or
xⴝ
or
sin x ⴝ 1
or
sin x ⴚ 1 ⴝ 0
So, an equation is: asin x ⴚ b (sin x ⴚ 1) ⴝ 0
1
2
This can be written as: (2 sin x ⴚ 1)(sin x ⴚ 1) ⴝ 0 or
2 sin2x ⴚ 3 sin x ⴙ 1 ⴝ 0
17. Determine the number of roots each equation has over the domain
0 ≤ x < 2␲, where 1 < a < b.
a) (a cos x ⫺ b)(b sin x ⫹ a) ⫽ 0
Solve the equation.
Either a cos x ⴚ b ⴝ 0
b
cos x ⴝ a
b) (b sin2x ⫺ 1)(a tan x ⫹ b) ⫽ 0
Solve the equation.
Either b sin2x ⴚ 1 ⴝ 0
b
But b>a, so a >1, and there
is no real solution
Or b sin x ⴙ a ⴝ 0
a
sin x ⴝ ⴚ
b
Since b>a, there is a root in the
two quadrants where the sine
ratio is negative.
So, there are 2 roots.
©P DO NOT COPY.
√1
sin x ⴝ —
Since b>1, 0<
√1
b
b
<1, and
there is a root in each quadrant.
Or a tan x ⴙ b ⴝ 0
b
tan x ⴝ ⴚ a
There is a root in the two
quadrants where the tangent
ratio is negative.
So, there is a total of 6 roots.
7.2 Solving Trigonometric Equations Algebraically—Solutions
13