What are the principal solutions of the equation $\tan x=-\frac{1}{\sqrt{3}}$?

The principal solutions of a trigonometric equation lie in the interval $[0,2\pi )$.Given:
$$\tan x=-\frac{1}{\sqrt{3}}$$
We know that $\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$.Since the value is negative, $x$ must lie in the second or fourth quadrant.In the second quadrant:
$$x=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$$
In the fourth quadrant:
$$x=2\pi -\frac{\pi }{6}=\frac{11\pi }{6}$$
Thus, the principal solutions are $\frac{5\pi }{6}$ and $\frac{11\pi }{6}$.