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

Given equation: $\cot x = -\sqrt{3}$.
We know that $\cot \frac{\pi}{6} = \sqrt{3}$.
Since $\cot x$ is negative in the second and fourth quadrants,we have:
$\cot (\pi - \frac{\pi}{6}) = -\cot \frac{\pi}{6} = -\sqrt{3} \Rightarrow \cot \frac{5\pi}{6} = -\sqrt{3}$.
$\cot (2\pi - \frac{\pi}{6}) = -\cot \frac{\pi}{6} = -\sqrt{3} \Rightarrow \cot \frac{11\pi}{6} = -\sqrt{3}$.
Thus,the principal solutions are $x = \frac{5\pi}{6}$ and $x = \frac{11\pi}{6}$.
For the general solution,we use the property that if $\cot x = \cot \alpha$,then $x = n\pi + \alpha$,where $n \in \mathbb{Z}$.
Taking the principal value $\alpha = \frac{5\pi}{6}$,the general solution is $x = n\pi + \frac{5\pi}{6}$,where $n \in \mathbb{Z}$.