Example 2 Define the function $f(x)$ on $\mathbf{R}$ that satisfies: $f(2+x)=2-f(x), f(x+3) \geqslant f(x)$. Try to find $f(x)$.
Solve: Since $f(x+3)=f[2+(x+1)]=2-f(x+1) \geqslant f(x)$, therefore
$$
f(x)+f(x+1) \leqslant 2,
$$

Also, $f(x)=2-f(2+x)$, substituting into $f(x+3) \geqslant f(x)$, we get
$f(x+3) \geqslant 2-f(x+2)$, i.e., $f(x+3)+f(x+2) \geqslant 2$, replacing $x$ with $x-2$, we get
$$
f(x)+f(x+1) \geqslant 2 \text {. }
$$

From (1) and (2), we have $f(x)+f(x+1)=2$, comparing with the known $f(x)=2-f(2+x)$, we get $f(x+1)=$ $f(2+x)$. Replacing $x$ with $x-1$, we get $f(x)=f(1+x)$, so
$f(x)=2-f(2+x)=2-f(1+x)=2-f(x)$, solving gives $f(x)=1$.