1. Solve the equation:

$$
\frac{\operatorname{cot} x-\operatorname{tan} x}{3 \sin x+\cos 2 x}=\operatorname{cot} 2 x
$$
1. Given the identities $\operatorname{tg} x=\frac{1}{\operatorname{ctg} x}$ and $\operatorname{tg} 2 x=\frac{2 \operatorname{tg} x}{1-\operatorname{tg}^{2} x}$, the given equation reduces to

$$
\frac{1-\operatorname{tg}^{2} x}{\operatorname{tg} x(3 \sin x+\cos 2 x)}=\frac{1-\operatorname{tg}^{2} x}{2 \operatorname{tg} x}
$$

Thus, either $1-\operatorname{tg}^{2} x=0$ or $3 \sin x+\cos 2 x=2$, under the condition that $\operatorname{tg} x$ is defined and $\operatorname{tg} x \neq 0$. In the first case, we get $\operatorname{tg} x= \pm 1$, which gives the solutions $x=\frac{\pi}{4}+\frac{k \pi}{2}, k \in \mathbb{Z}$. In the second case, we use the identity $\cos 2 x=1-2 \sin ^{2} x$, which reduces the given equation to $2 \sin ^{2} x-3 \sin x+1=0$; by substituting $t=\sin x$ and solving the resulting quadratic equation, $2 t^{2}-3 t+1=0$, we get $t_{1 / 2}=\frac{3 \pm \sqrt{9-8}}{4}=\frac{3 \pm 1}{4}$, i.e., $t=\sin x=1$ or $t=\sin x=\frac{1}{2}$. In the case
$\sin x=1$, $\operatorname{tg} x$ is not defined, so we are left with $\sin x=\frac{1}{2}$, which gives the solutions $x=\frac{\pi}{6}+2 k \pi$ or $x=\frac{5 \pi}{6}+2 k \pi, k \in \mathbb{Z}$. Therefore, all together we get:

$$
x \in\left\{\frac{\pi}{4}+\frac{k \pi}{2}: k \in \mathbb{Z}\right\} \cup\left\{\frac{\pi}{6}+2 k \pi: k \in \mathbb{Z}\right\} \cup\left\{\frac{5 \pi}{6}+2 k \pi: k \in \mathbb{Z}\right\}
$$