Find the principal solutions of the equation $\tan x=-\frac{1}{\sqrt{3}}.$

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We know that, $\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} .$

Thus, $\tan \left(\pi-\frac{\pi}{6}\right)=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}$

and $\quad \tan \left(2 \pi-\frac{\pi}{6}\right)=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}$

Thus $\quad \tan \frac{5 \pi}{6}=\tan \frac{11 \pi}{6}=-\frac{1}{\sqrt{3}}$

Therefore, principal solutions are $\frac{5 \pi}{6}$ and $\frac{11 \pi}{6}$ .

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  • [JEE MAIN 2019]