## Precalculus (6th Edition)

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)