11. The solution set of the equation $[\tan x]=2 \cos ^{2} x$ is
11. $\left\{x \left\lvert\, x=k \pi+\frac{\pi}{4}\right., k \in \mathbf{Z}\right\}$

Let $\tan x=y$, then from $2 \cos ^{2} x=\cos 2 x+1=\frac{1-\tan ^{2} x}{1+\tan ^{2} x}$, we know $[y]=\frac{1-y^{2}}{1+y^{2}}+1=\frac{2}{1+y^{2}}>0$, hence $[y] \geqslant 1$, so $y \geqslant 1$. From $y \geqslant 1$ we know $\frac{2}{1+y^{2}} \leqslant \frac{2}{1+1^{2}}=1$, thus $1 \leqslant[y] \leqslant 1$ which means $[y]=1$, hence $y=1$, i.e., $\tan x=1$.