Let $x = \arccos (\sin 2).$  Then
\begin{align*}
\cos x &= \sin 2 \\
&= \cos \left( \frac{\pi}{2} - 2 \right) \\
&= \cos \left( 2 - \frac{\pi}{2} \right).
\end{align*}Since $0 \le 2 - \frac{\pi}{2} \le \pi,$ $x = \boxed{2 - \frac{\pi}{2}}.$
Compute $\arccos (\sin 2).$  All functions are in radians.
{'answer': '2 - \\frac{\\pi}{2}', 'subject': 'Precalculus', 'level': 3, 'unique_id': 'train/precalculus/1182.json'}