The equation $\sin \left(x^{\operatorname{son} x}\right)=\cos \left(x^{\cos 2}\right)$ has the number of solutions in the closed interval $\left.1 \frac{\pi}{4}, \frac{\pi}{2}\right]$ is
(A) 0
(B) 2
(C) 8
(D) infinitely many
6. (A).

The equation is equivalent to $x^{40 \pi}+x^{\cos x}=\frac{\pi}{2}\left(\frac{\pi}{4} \leqslant x \leqslant \frac{\pi}{2}\right)$.
When $\frac{\pi}{4} \leqslant x \leqslant 1$, $x^{\sin x}>x$, so $x^{\sin x}+x^{\cos x}>2 x \geqslant \frac{\pi}{2}$, which is impossible. When $1 \leqslant x \leqslant \frac{\pi}{2}$, we have $x^{\sin x} \geqslant 1, x^{\cos x} \geqslant 1$, so $x^{\sin x}+x^{\cos x} \geqslant 2>\frac{\pi}{2}$, which is also impossible.