www.ck12.org
Chapter 3. Trigonometric Identities and Equations, Solution Key
3.3 Solving Trigonometric Equations
1. Answer:
• Because the problem deals with 2θ, the domain values must be doubled, making the domain 0 ≤ 2θ < 4π
• The reference angle is α = sin−1 0.6 = 0.6435
• 2θ = 0.6435, π − 0.6435, 2π + 0.6435, 3π − 0.6435
• 2θ = 0.6435, 2.4980, 6.9266, 8.7812
• The values for θ are needed so the above values must be divided by 2.
• θ = 0.3218, 1.2490, 3.4633, 4.3906
• The results can readily be checked by graphing the function. The four results are reasonable since they
are the only results indicated on the graph that satisfy sin 2θ = 0.6.
2.
cos2 x =
√
cos2 x =
1
16
r
1
16
1
cos x = ±
4
1
Then cos x =
4
−1 1
cos
=x
4
x = 1.3181 radians
or
cos x = −
1
4
1
cos−1 − = x
4
x = 1.8235 radians
• However, cos x is also positive in the fourth quadrant, so the other possible solution for cos x = 14 is
2π − 1.3181 = 4.9651 radians
• cos x is also negative in the third quadrant, so the other possible solution for cos x = − 41 is 2π − 1.8235 =
4.4597 radians
3.
tan2 x = 1
√
tan x = ± 1
tan x = ±1
41
3.3. Solving Trigonometric Equations
www.ck12.org
• So, tan x = 1 or tan x = −1.
• Therefore, x is all critical values corresponding with
π
4
5π 7π
within the interval. x = π4 , 3π
4 , 4 , 4
4. Use factoring by grouping.
2 sin x + 1 = 0
or
2 sin x = −1
1
sin x = −
2
7π 11π
x= ,
6 6
2 cos x − 1 = 0
2 cos x = 1
1
cos x =
2
π 5π
x= ,
3 3
5. You can factor this one like a quadratic.
sin2 x − 2 sin x − 3 = 0
(sin x − 3)(sin x + 1) = 0
sin x − 3 = 0
sin x + 1 = 0
sin x = 3
sin x = −1
3π
x=
2
or
x = sin−1 (3)
For this problem the only solution is
3π
2
because sine cannot be 3 (it is not in the range).
6.
tan2 x = 3 tan x
tan2 x − 3 tan x = 0
tan x(tan x − 3) = 0
tan x = 0
x = 0, π
42
or
tan x = 3
x = 1.25
www.ck12.org
Chapter 3. Trigonometric Identities and Equations, Solution Key
7. 2 sin2 4x − 3 cos 4x = 0
x
x
− 3 cos = 0
2 1 − cos2
4
4
x
2 x
2 − 2 cos − 3 cos = 0
4
4
x
2 x
2 cos + 3 cos − 2 = 0
4 4 x
x
2 cos − 1 cos + 2 = 0
4
4
.
&
x
x
2 cos − 1 = 0 or cos + 2 = 0
4
4
x
x
2 cos = 1
cos = −2
4
4
x 1
cos =
4 2
x π
5π
=
or
4 3
3
4π
20π
x=
or
3
3
20π
3
is eliminated as a solution because it is outside of the range and cos 4x = −2 will not generate any solutions
because −2 is outside of the range of cosine. Therefore, the only solution is 4π
3 .
8.
3 − 3 sin2 x = 8 sin x
3 − 3 sin2 x − 8 sin x = 0
3 sin2 x + 8 sin x − 3 = 0
(3 sin x − 1)(sin x + 3) = 0
3 sin x − 1 = 0
or
sin x + 3 = 0
3 sin x = 1
1
sin x =
sin x = −3
3
x = 0.3398 radians No solution exists
x = π − 0.3398 = 2.8018 radians
9. 2 sin x tan x = tan x + sec x
sin x
1
sin x
=
+
cos x cos x cos x
2 sin2 x sin x + 1
=
cos x
cos x
2
2 sin x = sin x + 1
2 sin x ·
2 sin2 x − sin x − 1 = 0
(2 sin x + 1)(sin x − 1) = 0
2 sin x + 1 = 0
2 sin x = −1
1
sin x = −
2
x=
or
sin x − 1 = 0
sin x = 1
7π 11π
,
6 6
43
3.3. Solving Trigonometric Equations
www.ck12.org
One of the solutions is not π2 , because tan x and sec x in the original equation are undefined for this value of x.
10.
2 cos2 x + 3 sin x − 3 = 0
2(1 − sin2 x) + 3 sin x − 3 = 0 Pythagorean Identity
2 − 2 sin2 x + 3 sin x − 3 = 0
− 2 sin2 x + 3 sin x − 1 = 0 Multiply by − 1
2 sin2 x − 3 sin x + 1 = 0
(2 sin x − 1)(sin x − 1) = 0
2 sin x − 1 = 0
2 sin x = 1
1
sin x =
2
π 5π
x= ,
6 6
or
sin x − 1 = 0
sin x = 1
x=
π
2
11. tan2 x + tan x − 2 = 0
p
12 − 4(1)(−2)
= tan x
2
√
−1 ± 1 + 8
= tan x
2
−1 ± 3
= tan x
2
tan x = −2 or 1
tan x = 1 when x = π4 , in the interval − π2 , π2 tan x = −2 when x = −1.107 rad
12. 5 cos2 θ − 6 sin θ = 0 over the interval [0, 2π].
5 1 − sin2 x − 6 sin x = 0
−1 ±
−5 sin2 x − 6 sin x + 5 = 0
5 sin2 x + 6 sin x − 5 = 0
p
−6 ± 62 − 4(5)(−5)
= sin x
2(5)
√
−6 ± 36 + 100
= sin x
10 √
−6 ± 136
= sin x
10 √
−6 ± 2 34
= sin x
10 √
−3 ± 34
= sin x
5
√ √ x = sin−1 −3+ 5 34 or sin−1 −3− 5 34 x = 0.6018 rad or 2.5398 rad from the first expression, the second
expression will not yield any answers because it is out the the range of sine.
44