5. Let $f(x)$ be a function defined on $\mathbf{R}$ with a period of 2, which is even, strictly decreasing on the interval $[0,1]$, and satisfies $f(\pi)=1, f(2 \pi)=2$. Then the solution set of the inequality system $\left\{\begin{array}{l}1 \leqslant x \leqslant 2, \\ 1 \leqslant f(x) \leqslant 2\end{array}\right.$ is $\qquad$.
Notice that $f(x)$ is monotonically increasing on the interval $[1,2]$, $1<\pi-2<8-2\pi<2$, and $f(\pi-2)=f(\pi)=1, f(8-2\pi)=f(2\pi-8)=f(2\pi)=2$, so the solution set of the inequality system is $[\pi-2,8-2\pi]$.