Part II: The Answers
174
Answers
101–200
135.
π
3
, π, 5
3
π
Factor the equation:
2
1 0
2
1
1 0
2
cos
cos
( cos
)(cos
)
x
x
x
x
+
− =
−
+ =
Setting the first factor equal to zero gives you 2 cos x – 1 = 0 so that cos x = 1
2
,
which has the solutions x = π
3
and 5
3
π . Setting the second factor equal to zero
gives you cos x + 1 = 0 so that cos x = –1, which has the solution x = π.
136.
π
4
, 3
4
π , 5
4
π , 7
4
π
To solve the equation tan x = 1, you must find the solutions to two equations: tan x = 1
and tan x = –1. For the equation tan x = 1, you have the solutions x = π
4
and 5
4
π . For the
equation tan x = –1, you have the solutions x = 3
4
π and 7
4
π .
137.
7
6
π , 11
6
π
Factor the equation 2 sin
2 
x – 5 sin x – 3 = 0 to get
( sin
)(sin
)
2
1
3 0
x
x
+
− =
Setting the first factor equal to zero gives you 2 sin x + 1 = 0 so that sin x = − 1
2
,
which has the solutions x = 7
6
π and 11
6
π . Setting the second factor equal to zero
gives you sin x – 3 = 0 so that sin x = 3. Because 3 is outside the range of the sine function, this equation from the second factor has no solution.
138.
π
2
, 3
2
π
Begin by making one side of the equation equal to zero. Then use the identity
cot
cos
sin
x
x
x
=
and factor:
cos
cot
cos
cot
cos
cos
sin
cos
sin
x
x
x
x
x
x
x
x
x
=
−
=
−
=
−
(
) =
0
0
1
1
0
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 1
1
0
−
=
sin x
so that sin x = 1,
which has the solution x = π
2
.