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

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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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