Given $f(x)$ is a function defined on the set of real numbers with a period of 2, and it is an even function. It is known that when $x \in [2,3]$, $f(x) = x$. Find the analytical expression of $f(x)$ when $x \in [-2,0]$.
Analysis From $T=2$, we can derive the analytical expressions of $f(x)$ when $x \in[-2,-1]$ and $x \in[0,1]$. Then, by the odd-even properties, we can obtain the analytical expressions of $f(x)$ when $x \in[-2,0]$.

Solution Since the function $f(x)$ is a periodic function with a period of $T=2$, we have $f(x+2)=f(x)$. When $x \in[-2,-1]$, $x+4 \in[2,3]$, so $f(x+4)=x+4$, then $f(x)=f(x+4)=x+4$. When $x \in[0,1]$, $x+2 \in[2,3]$, so $f(x+2)=x+2$, then $f(x)=f(x+2)=x+2$. Since $f(x)$ is an even function, we have $f(-x)=f(x)$. When $x \in[-1,0)$, $-x \in(0,1]$, then $f(x)=f(-x)=-x+2$. Therefore, $f(x)=\left\{\begin{array}{l}3-(-x-1)=x+4(x \in[-2,-1]) \\ 3-(x+1)=-x+2(x \in(-1,0))\end{array}=3-|x+1|(x \in[-2\right.$, $0])$.