Solve $2 \cos ^{2} x+3 \sin x=0$
Solve $2 \cos ^{2} x+3 \sin x=0$
The equation can be written as
$2\left(1-\sin ^{2} x\right)+3 \sin x=0$
or $2 \sin ^{2} x-3 \sin x-2=0$
or $(2 \sin x+1)(\sin x-2)=0$
Hence $\sin x=-\frac{1}{2} \quad$ or $\quad \sin x=2$
But $\sin x=2$ is not possible (Why?)
Therefore $\sin x=-\frac{1}{2}=\sin \frac{7 \pi}{6}$
Hence, the solution is given by
$x=n \pi+(-1)^{n} \frac{7 \pi}{6}, \text { where } n \in Z.$
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