**Solution:** The equation can be written as

2(1− sin2 x) + 3sin x = 0

or 2 sin2 x − 3sin x − 2 = 0

or (2sinx +1) (sinx − 2) =0

Hence sin x = -1/2 or sin x = 2

But sin x = 2 is not possible

Therefore sin x = -1/2 = sin 7π/6

Hence, the solution is given by

**x = nπ + (-1)**^{n}7π/6, where n ∈ Z.