## Precalculus (6th Edition)

$y =\displaystyle \frac{5\pi}{6}$
$y=\cos^{-1}x =\arccos x$ Domain:$[-1, 1]$ Range: $[0, \pi]$ ----------- $y$ is the number from $[0, \pi]$ such that $\displaystyle \cos y=-\frac{\sqrt{3}}{2}.$ Taking $\displaystyle \frac{\pi}{6}$ as a reference angle,$\displaystyle \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$, we have (in quadrant II) $\displaystyle \cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}\qquad$and$\displaystyle \quad \frac{5\pi}{6}\in [0, \pi]$ , so $y =\displaystyle \frac{5\pi}{6}$