Answers and Explanations 175
Answers
101–200
139.
π
12
, 5
2
π
1
, 13
12
π , 17
12
π
Begin by making the substitution 2x = y. Because 0 ≤ x ≤ 2π, it follows that 0 ≤ 2x ≤ 4π
and that 0 ≤ y ≤ 4π.
Solving the equation sin( )
y = 1
2
gives you the solutions y = π
6
, 5
6
π , 13
6
π , and 17
6
π in the
interval [0, 4π].
Next, take each solution, set it equal to 2x, and solve for x: 2
6
x = π so that x = π
12
;
2
5
6
x = π so that x = 1
5
2
π ; 2
13
6
x = π so that x = 13
12
π ; and 2
17
6
x = π so that x = 17
12
π .
140.
π
6
, π
2
, 5
6
π , 3
2
π
Begin by making one side of the equation zero. Then use the identity
sin 2x = 2 sin x cos x and factor:
sin
cos
sin
cos
sin cos
cos
cos ( sin
)
2
2
0
2
0
2
1 0
x
x
x
x
x
x
x
x
x
=
−
=
−
=
− =
Setting the first factor equal to zero gives you cos x = 0, which has the solutions x = π
2
and 3
2
π . Setting the second factor equal to zero gives you 2 sin x – 1 = 0 so that
sin x = 1
2
, which has the solutions x = π
6
and 5
6
π .
141.
π
2
, 3
2
π
Begin by using the identity sin
s in cos
2
2
x
x
x
=
and then factor:
2
2
0
2
2
0
2
1
0
cos
sin
cos
sin cos
cos ( sin )
x
x
x
x
x
x
x
+
=
+
=
+
=
Set each factor equal to zero and solve for x. The equation cos x = 0 has the
solutions x = π
2
and 3
2
π , and the equation 1 + sin x = 0 gives you sin x = –1, which
has the solution x = 3
2
π .
142.
2
3
π , π , 4
3
π
Use the identity cos 2x = 2 cos
2 
x – 1, make one side of the equation zero, and factor:
2
2
3
2 2
1
3
2
3
1 0
2
1
2
2
+
=−
+
− = −
+
+ =
+
cos
c os
cos
c os
cos
cos
( cos
)(c
x
x
x
x
x
x
x
o os
)
x + =
1 0
Set each factor equal to zero and solve for x. Setting the first factor equal to zero