#### Answer

A

#### Work Step by Step

A. From the definition of arcsin,
x is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ for which $\displaystyle \sin x=\frac{\sqrt{2}}{2}.$
$\displaystyle \frac{\pi}{4}\in [-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ and $\displaystyle \sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$ so
$\displaystyle \frac{\pi}{4}$ is the solution.
B. From the definition of arccos,
x is the number from $[0, \pi]$ for which $\displaystyle \cos x=-\frac{\sqrt{2}}{2}.$
$\displaystyle \frac{\pi}{4}\in [0, \pi] $and $\displaystyle \cos\frac{\pi}{4}=+\frac{\sqrt{2}}{2}$ so
$\displaystyle \frac{\pi}{4}$ is NOT the solution ($\displaystyle \frac{3\pi}{4}$ is).
C. 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}{4}\in (-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}) $and $\displaystyle \tan\frac{\pi}{4}=$1 so
$\displaystyle \frac{\pi}{4}$ is NOT the solution ($\displaystyle \frac{\pi}{6}$ is)