1. Solve the equation:

$$
\sqrt{17-7 \sin 2 x}=3 \cos x-5 \sin x
$$
1. From $\sin 2 x \leqslant 1$ we have $17-7 \sin 2 x>0$ for all values of $x$, so the root on the left side is always defined. Therefore, before squaring the equation, we only need to set the condition $3 \cos x-5 \sin x>0$. After squaring, we get $17-7 \sin 2 x=9 \cos ^{2} x+$ $25 \sin ^{2} x-30 \sin x \cos x$, i.e., substituting $17=17 \sin ^{2} x+17 \cos ^{2} x$ and moving everything to the left side,

$$
8 \cos ^{2} x-8 \sin ^{2} x+30 \sin x \cos x-7 \sin 2 x=0
$$

From the identities $\cos ^{2} x-\sin ^{2} x=\cos 2 x$ and $2 \sin x \cos x=\sin 2 x$, we see that the obtained equation is equivalent to $8 \cos 2 x+8 \sin 2 x=0$. This can be further transformed as

$0=8 \sqrt{2}\left(\frac{\sqrt{2}}{2} \cos 2 x+\frac{\sqrt{2}}{2} \sin 2 x\right)=8 \sqrt{2}\left(\sin \frac{\pi}{4} \cos 2 x+\cos \frac{\pi}{4} \sin 2 x\right)=8 \sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)$,

from which it follows that $2 x+\frac{\pi}{4}=k \pi, k \in \mathbb{Z}$. Thus, the solutions to the last equation are numbers of the form $x=-\frac{\pi}{8}+\frac{k \pi}{2}, k \in \mathbb{Z}$. These can be divided into four classes (where in each class we have a period of $2 \pi): x=-\frac{\pi}{8}+2 k \pi, x=\frac{3 \pi}{8}+2 k \pi, x=\frac{7 \pi}{8}+2 k \pi, x=\frac{11 \pi}{8}+2 k \pi$, $k \in \mathbb{Z}$. The solutions from the second and third classes do not satisfy the condition $3 \cos x-5 \sin x>0$ (we have $\cos \frac{3 \pi}{8}\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$, so $3 \cos \frac{3 \pi}{8}-5 \sin \frac{3 \pi}{8}0$, and $\sin \left(-\frac{\pi}{8}\right)0$; also, $\cos \frac{11 \pi}{8}>\cos \frac{5 \pi}{4}=-\frac{\sqrt{2}}{2}$ and $\sin \frac{11 \pi}{8}\sqrt{2}>0$), so only they form the solutions to the original equation. In other words, the answer is:

$$
x \in\left\{-\frac{\pi}{8}+2 k \pi, \frac{11 \pi}{8}+2 k \pi: k \in \mathbb{Z}\right\}
$$