Answer
C
Work Step by Step
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.
