Algebra and Trigonometry 10th Edition

$arccos(-\frac{\sqrt 2}{2})=\frac{3\pi}{4}$
The range of $arccos~x$ is $[0,\pi]$. So, $arccos(-\frac{\sqrt 2}{2})=\frac{3\pi}{4}$, because $cos(\frac{3\pi}{4})=-\frac{\sqrt 2}{2}$ and $0\leq\frac{3\pi}{4}\leq\pi$