Part II: The Answers
184
Answers
101–200
3
0963
0 321
x
x
=
=
.
.
3
2
0 963
2
3
0 321
x
x
=
−
=
−
π
π
.
.
3
2
0 963
2
3
0 321
x
x
=
+
=
+
π
π
.
.
3
4
0 963
4
3
0 321
x
x
=
−
=
−
π
π
.
.
3
4
0 963
4
3
0 321
x
x
=
+
=
+
π
π
.
.
3
6
0 963
2
0321
x
x
=
−
=
−
π
π
.
.
Therefore, the solutions are x = 0.321, 2
3
0 321
π − .
, 2
3
0 321
π + .
, 4
3
0 321
π − .
, 4
3
0 321
π + .
, and
2π – 0.321.
165.
x = π + 0.887, 2π – 0.887
To solve 2 sin
2 
x + 8 sin x + 5 = 0 for x over the interval [0, 2π], first use the quadratic
formula:
sin
( )( )
( )
( )
x =
− ±
−
= − ±
= − ±
8
8 4 2 5
2 2
8
24
2 2
8
4
24
4
2
Simplifying gives you − +
≈ −
2
24
4
0 775
.
and − −
≈ −
2
24
4
3 225
.
.
You now need to find solutions to sin x = –0.775 and sin x = –3.225. Notice that
sin x = –3.225 has no solutions because –3.225 is outside the range of the sine function.
To solve sin x = –0.775, take the inverse sine of both sides:
sin
.
sin ( .
)
.
x
x
= −
=
−
≈ −
−
0 775
0 775
0 887
1
Note that this solution isn’t in the desired interval, [0, 2π]. The solutions in
the given interval belong to Quadrants III and IV, respectively, because in
those quadrants, the sine function has negative values; those solutions are
x = π + 0.887 and x = 2π – 0.887.