Solve $2 \cos^{2} x + 3 \sin x = 0$.

(N/A) The given equation is $2 \cos^{2} x + 3 \sin x = 0$.
Using the identity $\cos^{2} x = 1 - \sin^{2} x$,we get:
$2(1 - \sin^{2} x) + 3 \sin x = 0$
$2 - 2 \sin^{2} x + 3 \sin x = 0$
$2 \sin^{2} x - 3 \sin x - 2 = 0$
Factoring the quadratic equation:
$(2 \sin x + 1)(\sin x - 2) = 0$
This gives two cases:
$1) \sin x = -\frac{1}{2}$
$2) \sin x = 2$
Since the range of $\sin x$ is $[-1, 1]$,$\sin x = 2$ is impossible.
For $\sin x = -\frac{1}{2}$,we know $\sin x = \sin(-\frac{\pi}{6})$.
The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^{n}\alpha$.
Thus,$x = n\pi + (-1)^{n}(-\frac{\pi}{6})$,where $n \in \mathbb{Z}$.