Find the principal and general solutions of the equation $cot x=-sqrt{3}$

$\cot x=-\sqrt{3}$

It is known that $\cot \frac{\pi}{6}=\sqrt{3}$

$\therefore \cot \left(\pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-\sqrt{3}$ and $\cot \left(2 \pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-\sqrt{3}$

i.e., $\cot \frac{5 \pi}{6}=-\sqrt{3}$ and $\cot \frac{11 \pi}{6}=-\sqrt{3}$

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

Now, $\cot x=\cot \frac{5 \pi}{6}$

$\Rightarrow \tan x=\tan \frac{5 \pi}{6} \quad\left[\cot x=\frac{1}{\tan x}\right]$

$\Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{5 \pi}{6}$, where $\mathrm{n} \in \mathrm{Z}$

Therefore, the general solution is $x=n \pi+\frac{5 \pi}{6}$, where $n \in Z$