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The principal solution of $\cot x=\sqrt{3}$ are
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$\frac{\pi}{6}, \frac{7 \pi}{6}$
$\begin{aligned} & \cot x=\sqrt{3} \quad \Rightarrow \quad \tan x=\frac{1}{\sqrt{3}} \\ & \therefore \frac{1}{\sqrt{3}}=\tan \left(\pi+\frac{\pi}{6}\right)=\tan \left(\frac{\pi}{6}\right) \\ & \Rightarrow \frac{1}{\sqrt{3}}=\tan \left(\frac{7 \pi}{6}\right)=\tan \left(\frac{\pi}{6}\right)\end{aligned}$
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