3. The solution to the inequality about $x$
$$
\left|\log _{\cos x} \sin x\right|>\left|\log _{\sin x} \cos x\right|
$$

is $\qquad$
$\begin{array}{l}\text { 3. }\left(2 k \pi, 2 k \pi+\frac{\pi}{4}\right)(k \in \mathbf{Z}) . \\ \left|\log _{\cos x} \sin x\right|>\left|\log _{\sin x} \cos x\right| \\ \Leftrightarrow\left\{\left|\frac{\lg \sin x}{\lg \cos x}\right|>\left|\frac{\lg \cos x}{\lg \sin x}\right|\right. \\ \Leftrightarrow 0<\sin x<\cos x<1 \\ \Leftrightarrow 0<\tan x<1 \\ \Leftrightarrow 2 k \pi<x<2 k \pi+\frac{\pi}{4}(k \in \mathbf{Z}) .\end{array}$