## Precalculus (10th Edition)

$\frac{\pi}{6}$
$\sec^{-1}{\frac{2\sqrt3}{3}}=\frac{\pi}{6}$, because $\sec{\frac{\pi}{6}}=\frac{2\sqrt3}{3}$ and $\frac{\pi}{6}$ is in the range of $\sec^{-1}{x}$, which is $[0,\pi]$ \ $\left\{\frac{\pi}{2}\right\}.$