Solve the equation $2-\log _{\sin x} \cos x=\log _{\cos x} \sin x$.
Let's start by writing down the domain of the equation: $\sin x>0, \cos x>0$. Convert the logarithm on the right side to the base $\sin x: 2-\log _{\sin x} \cos x=1 / \log _{\sin x} \cos x$. After transformations, we get:

$$
\left(\log _{\sin x} \cos x-1\right)^{2}=0 \text { or } \log _{\sin x} \cos x=1
$$

After exponentiation, we will have:

$$
\cos x=\sin x, x=\pi / 4+k \pi \text {. }
$$

Among the obtained values, only solutions of the form $x=\pi / 4+2 k \pi, k \in Z$ satisfy the domain.

## Answer

$x=\pi / 4+2 k \pi, k \in Z$.