$5.57 \sin ^{3} x(1+\operatorname{ctg} x)+\cos ^{3} x(1+\operatorname{tg} x)=2 \sqrt{\sin x \cos x}$.
5.57 We have $\sin x \cos x>0$, but for $\sin x>0, \cos x>0$. Transform the left side of the equation:

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

## Then we get

$\sin x+\cos x-2 \sqrt{\sin x \cos x}=0 ;(\sqrt{\sin x}-\sqrt{\cos x})^{2}=0 ;$

$\sin x=\cos x, \operatorname{tg} x=1 ; x=\frac{\pi}{4}+2 \pi k, \operatorname{so}$ since $\sin x>0, \cos x>0$.

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