8.047. $\sin ^{3} z \cos z-\sin z \cos ^{3} z=\frac{\sqrt{2}}{8}$.
## Solution.

From the condition

$-\sin z \cos z\left(\cos ^{2} z-\sin ^{2} z\right)=\frac{\sqrt{2}}{8} \Leftrightarrow 2 \sin z \cos z\left(\cos ^{2} z-\sin ^{2} z\right)=-\frac{\sqrt{2}}{4} \Leftrightarrow$ $\Leftrightarrow \sin 2 z \cos 2 z=-\frac{\sqrt{2}}{4} \Leftrightarrow 2 \sin 2 z \cos 2 z=-\frac{\sqrt{2}}{2}, \sin 4 z=-\frac{\sqrt{2}}{2}$,

from which $4 z=(-1)^{k+1} \frac{\pi}{4}+\pi k, \quad z=(-1)^{k+1} \frac{\pi}{16}+\frac{\pi k}{4}, \quad k \in Z$.

Answer: $\quad z=(-1)^{k+1} \frac{\pi}{16}+\frac{\pi k}{4}, k \in Z$.