Find the principal value of the equation $\sqrt{3} \sec \theta = 2$
$\sqrt{3} \sec \theta = 2$ $\implies$ $\sec \theta = \dfrac{2}{\sqrt{3}}$ $\implies$ $\cos \theta = \dfrac{\sqrt{3}}{2} > 0$
$\therefore \;$ $\theta$ lies in the first or fourth quadrants.
Now, $\cos \theta = \dfrac{\sqrt{3}}{2} = \cos \left(\dfrac{\pi}{6}\right)$
$\implies$ $\theta = \dfrac{\pi}{6}$, $\;$ $\theta = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6}$
Principal value of $\cos \theta$ is in $\left[0, \pi\right]$ $\;$ i.e. $\;$ in first or second quadrants.
$\because \;$ $\dfrac{11 \pi}{6} \notin \left[0, \pi\right]$,
$\therefore \;$ Principal value of the given equation is $\dfrac{\pi}{6}$