## Precalculus (6th Edition)

$y=\displaystyle \frac{\pi}{6}$
$y=\sec^{-1}x =$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 -------- $y$ 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, $y=\displaystyle \frac{\pi}{6}$