$8.473 \sin ^{5} x+\cos ^{5} x=2-\sin ^{4} x$
Solution. $\sin ^{5} x \leq \sin ^{2} x, \cos ^{5} x \leq \cos ^{2} x \Rightarrow \sin ^{5} x+\cos ^{5} x \leq 1$. But $2-\sin ^{4} x \geq 1$ for all $x$. Therefore, the original equation is equivalent to $\left\{\begin{array}{l}\sin ^{5} x=\sin ^{2} x \\ \cos ^{5} x=\cos ^{2} x \\ \sin ^{4} x=1 .\end{array}\right.$

Answer: $\quad x=\frac{\pi}{2}(1+4 k)$.