8.41 Given the equation $\arccos \frac{4}{5}-\arccos \left(-\frac{4}{5}\right)=\arcsin x$, then
(A) $x=\frac{24}{25}$.
(B) $x=-\frac{24}{25}$.
(C) $x=0$.
(D) Such an $x$ does not exist.
(China High School Mathematics Competition, 1985)
［Solution］The original equation can be rewritten as $2 \arccos \frac{4}{5}-\pi=\arcsin x$. Since
$$
\begin{array}{c}
0<\arccos \frac{4}{5}<\arccos \frac{\sqrt{2}}{2}=\frac{\pi}{4}, \\
0<2 \arccos \frac{4}{5}<\frac{\pi}{2} .
\end{array}
$$

and $-\frac{\pi}{2} \leqslant \arcsin x \leqslant \frac{\pi}{2}$, the given equation has no solution. Therefore, the answer is $(D)$.