Let
\[f(x) = (\arccos x)^2 \cdot (\arcsin x)^2.\]
Find the range of $f(x).$ All functions are in radians.
We take cases.
First, suppose $-1 \le x \le 1.$ Then $\arccos x$ and $\arcsin x$ are both defined. Since $-1 \le x \le 1,$
\[0 \le \arccos x \le \frac{\pi}{2}\]and
\[-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2},\]so
\[0 \le \arccos x \cdot \arcsin x \le \frac{\pi^2}{4},\]which means
\[f(x) = (\arccos x)^2 \cdot (\arcsin x)^2 \le \left( \frac{\pi^2}{4} \right)^2 = \frac{\pi^4}{16}.\]
Next, suppose $x > 1.$ We use the identity
\[\arcsin x = \frac{\pi}{2} - \arccos x,\]so $f(x) = (\arccos x)^4.$ Then $\arccos x$ is defined, by $0 \le \arccos x \le \pi,$ so $(\arccos x)^4$ is defined, by $0 \le (\arccos x)^4 \le \pi^4,$ so $0 \le f(x) \le \pi^4.$
However, once $x \ge 1,$ $\arcsin x$ is not defined, so we take the right endpoint out of the range. Thus, the range is $\boxed{\left[ 0, \frac{\pi^4}{16} \right]}.$