8.310. $\sin ^{3} x(1-\operatorname{ctg} x)+\cos ^{3} x(1-\operatorname{tg} x)=1.5 \cos 2 x$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0 \\ \cos x \neq 0 .\end{array}\right.$

## Write the equation in the form

$$
\begin{aligned}
& \sin ^{3} x\left(1-\frac{\cos x}{\sin x}\right)+\cos ^{3} x\left(1-\frac{\sin x}{\cos x}\right)-1.5 \cos 2 x=0 \Leftrightarrow \\
& \Leftrightarrow \sin ^{2} x(\sin x-\cos x)-\cos ^{2} x(\sin x-\cos x)-1.5 \cos 2 x=0 \Leftrightarrow
\end{aligned}
$$

$\Leftrightarrow(\cos x-\sin x)\left(\cos ^{2} x-\sin ^{2} x\right)-1.5 \cos 2 x=0 \Leftrightarrow$

$\Leftrightarrow(\cos x-\sin x) \cos 2 x-1.5 \cos 2 x=0 \Leftrightarrow \cos 2 x(\cos x-\sin x-1.5)=0$.

## From this

$\cos 2 x=0, 2 x=\frac{\pi}{2}+\pi k, x=\frac{\pi}{4}+\frac{\pi k}{2}=\frac{\pi}{4}(2 k+1), k \in Z$;

$\cos x-\sin x-1.5 \neq 0$.

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