* Let $[\tan x]$ denote the greatest integer not exceeding the real number $\tan x$. Then the solution to the equation $[\tan x]=2 \cos ^{2} x$ is $\qquad$
Given $0 \leqslant 2 \cos ^{2} x \leqslant 2$, hence $[\tan x]$ can only take the values 0, 1, 2. When $[\tan x]=0$, $\cos x=0$, at this point $\tan x$ is undefined. When $[\tan x]=2$, $\cos ^{2} x=1$, at this point $\tan x=0$, which is impossible. When $[\tan x]=1$, $\cos x= \pm \frac{\sqrt{2}}{2}$, i.e., $x= \pm \frac{\pi}{4}+k \pi, k \in \mathbf{Z}$, note that $[\tan x]=$ 1, so it can only be $x=\frac{\pi}{4}+k \pi, k \in \mathbf{Z}$.