#### Answer

$y=\displaystyle \frac{\pi}{6}$

#### Work Step by Step

$y=\sec^{-1}x =$arcsec$x$
Domain: $(-\infty, -1]\cup[1, \infty)$
Range: $[0,\displaystyle \frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$
Quadrants (unit circle): I and II
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$y$ is the number from $[0,\displaystyle \frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$
such that $\displaystyle \sec y=\frac{2\sqrt{3}}{3}$
$(\displaystyle \cos y=\frac{3}{2\sqrt{3}}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2})$
In quadrant I, $\displaystyle \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$, that is,
$\displaystyle \sec\frac{\pi}{6}=\frac{2\sqrt{3}}{3}, $
and
$\displaystyle \frac{\pi}{6}\in[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$ .
So,
$y=\displaystyle \frac{\pi}{6}$