$5.52 \operatorname{ctg} x-\operatorname{tg} x=\sin x+\cos x$.

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$5.52 \operatorname{ctg} x-\operatorname{tg} x=\sin x+\cos x$.
5.52 Here $\sin x \neq 0, \cos x \neq 0$. We have:

$$
\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}=\sin x+\cos x
$$

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

$(\sin x+\cos x)(\cos x-\sin x-\sin x \cos x)=0 ;$

1) $\sin x+\cos x=0 ; \operatorname{tg} x=-1 ; x_{1}=\frac{\pi}{4}(4 k-1)$;
2) $\cos x-\sin x-\sin x \cos x=0$. Let $\cos x-\sin x=y$; then

$(\cos x-\sin x)^{2}=y^{2}$, or $1-2 \sin x \cos y=y^{2}$,

i.e., $\sin x \cos x=\frac{1-y^{2}}{2}$. We solve the obtained quadratic equation:

$y-\frac{1-y^{2}}{2}=0 ; y^{2}+2 y-1=0 ; y=-1 \pm \sqrt{2}$;

$\cos x-\sin x \neq-1-\sqrt{2}$, since $-1-\sqrt{2}<-2$ (recall that $|\cos x-\sin x| \leqslant \sqrt{2}$ ). Therefore,

$$
\cos x-\sin x=-1+\sqrt{2} ; \cos x-\cos \left(\frac{\pi}{2}-x\right)=-1+\sqrt{2}
$$

$$
\begin{aligned}
& 2 \sin \frac{\pi}{4} \sin \left(\frac{\pi}{4}-x\right)=-1+\sqrt{2} ; \sin \left(\frac{\pi}{4}-x\right)=\frac{\sqrt{2}-1}{\sqrt{2}} ; \\
& x_{2}=(-1)^{k} \arcsin \frac{\sqrt{2}-2}{2}+\frac{\pi}{4}(4 k+1)
\end{aligned}
$$

Answer: $\quad x_{1}=\frac{\pi}{4}(4 k-1), x_{2}=(-1)^{k} \arcsin \frac{\sqrt{2}-2}{2}+\frac{\pi}{4}(4 k+1), k \in Z$.