Find the solution of $\sin x=-\frac{\sqrt{3}}{2}$

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We have $\sin x=-\frac{\sqrt{3}}{2}$

$=-\sin \frac{\pi}{3}=\sin \left(\pi+\frac{\pi}{3}\right)$

$=\sin \frac{4 \pi}{3}$

Hence $\sin x=\sin \frac{4 \pi}{3},$ which gives

$x=n \pi+(-1)^{n} \frac{4 \pi}{3}, \text { where } n \in Z$

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