Answer
$arccos[cos(-\frac{3\pi}{2})]=\frac{\pi}{2}$
Work Step by Step
$-\frac{3\pi}{2}$ does not lie in the range of $arccos~x$: ($0\leq x\leq \pi$). But, we know that the period of $cos~x=2\pi$ and that $-\frac{3\pi}{2}=\frac{\pi}{2}-2\pi$. So:
$cos(-\frac{3\pi}{2})=cos(\frac{\pi}{2})$. Now, $\frac{\pi}{2}$ lies in the range of $arccos~x$.
$arccos[cos(-\frac{3\pi}{2})]=\frac{\pi}{2}$
