## Precalculus (6th Edition)

A. From the definition of arctan, x is the number from $(-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2})$ for which $\displaystyle \tan x=\frac{\sqrt{3}}{3}$ $-\displaystyle \frac{\pi}{6}\in (-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2})$ and $\tan$($-\displaystyle \frac{\pi}{6}$)$=-\displaystyle \frac{\sqrt{3}}{3}$ so $-\displaystyle \frac{\pi}{6}$ is NOT the solution ($\displaystyle \frac{\pi}{6}$ is). B. From the definition of arccos, x is the number from $[0, \pi]$ for which $\displaystyle \cos x=-\frac{1}{2}.$ $-\displaystyle \frac{\pi}{6}\not\in [0, \pi]$ so $-\displaystyle \frac{\pi}{6}$ is not the solution. C. From the definition of arcsin, x is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ for which $\displaystyle \sin x=-\frac{1}{2}.$ $-\displaystyle \frac{\pi}{6}\in [-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ and $\displaystyle \sin(-\frac{\pi}{6})=-\frac{1}{2}$ so $-\displaystyle \frac{\pi}{6}$ is the solution.