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Problem
Problem
Find the number of solutions of the equation
2
sin
3
x
+
6
sin
2
x
−
sin
x
−
3
=
0
2\sin^3 x + 6\sin^2 x - \sin x - 3 = 0
2
sin
3
x
+
6
sin
2
x
−
sin
x
−
3
=
0
in
(
0
,
2
π
)
(0, 2\pi)
(
0
,
2
π
)
.
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