One of principal solution of $\sqrt 3 \sec x = - 2$ is equal to

$\begin{aligned} \sqrt{3} \sec x & =-2 \\ \sec x & =\frac{-2}{\sqrt{3}} \Rightarrow \cos x=\frac{-\sqrt{3}}{2} \end{aligned}$

$\cos \theta$ is negative in 2nd and 3rd quadrant

$\begin{aligned} \therefore \cos x & =\frac{-\sqrt{3}}{2} \\ & =\pi-\frac{\pi}{6}, \pi+\pi / 6 \\ & =\frac{6 \pi-\pi}{6}, \frac{6 \pi+\pi}{6} \\ x & =\{5 \pi / 6,7 \pi / 6\} \end{aligned}$