Example 16 (from the 2nd American Mathematical Invitational Competition) The function $f(x)$ is defined on the real number domain and satisfies the following conditions: for any real number $x$, $f(2+x)=f(2-x)$, and $f(7+x)=f(7-x)$. If $x=0$ is a root of the equation $f(x)=0$, how many roots should the equation $f(x)=0$ have at least in the interval $-1000 \leqslant x \leqslant 1000$?
From the function $f(x)$ for any real number $x$, we have $f(2+x)=f(2-x), f(7+x)=f(7-x)$, which indicates that $f(x)$ is a generalized (II) or (II) type even function, with $2|2-7|=10$ or $|(2+2)-(7+7)|=10$ being one of its positive periods (Conclusion 5(1) or Conclusion 7(1)).

Furthermore, from $0=f(0)=f(2-2)=f(2+2)=f(4)=f(7-3)=f(7+3)=f(10)$, we know that in the intervals $-10(k+1) \leqslant x<-10 k$ and $10 k<x \leqslant 10(k+1) (k=0,1,2, \cdots, 99)$, $f(x)=0$ has at least 2 roots, totaling 400 roots. Additionally, since $f(0)=0$, $f(x)=0$ has at least 401 roots in the interval $-1000 \leqslant x \leqslant 1000$.