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