13. Given the function $f(x)$ $=\sin 2 x \cdot \tan x+\cos \left(2 x-\frac{\pi}{3}\right)-1$, find the intervals where the function $f(x)$ is monotonically decreasing.
$$
\begin{array}{l}
f(x)=2 \sin ^{2} x+\frac{1}{2} \cos 2 x+\frac{\sqrt{3}}{2} \sin 2 x-1=\frac{\sqrt{3}}{2} \sin 2 x-\frac{1}{2} \cos 2 x=\sin \left(2 x-\frac{\pi}{6}\right), \\
2 x-\frac{\pi}{6} \in\left[2 k \pi+\frac{\pi}{2}, 2 k \pi+\frac{3 \pi}{2}\right] \Rightarrow x \in\left[k \pi+\frac{\pi}{3}, k \pi+\frac{5 \pi}{6}\right](k \in \mathbf{Z}),
\end{array}
$$

Notice that the domain of $f(x)$ is $\left(k \pi-\frac{\pi}{2}, k \pi+\frac{\pi}{2}\right)(k \in \mathbf{Z})$,
so the decreasing interval of $f(x)$ is $\left[k \pi+\frac{\pi}{3}, k \pi+\frac{\pi}{2}\right),\left(k \pi+\frac{\pi}{2}, k \pi+\frac{5 \pi}{6}\right](k \in \mathbf{Z})$.