Example 9 Given that the function $f(x)$ defined on $\mathbf{R}$ is an odd function, and is increasing on $[0,+\infty)$. For any $\theta \in \mathbf{R}$, find the real number $m$ such that $f(\cos 2 \theta-3)+f(4 m-$ $2 m \cos \theta)>0$ always holds.
Strategy: Utilize the properties of the function, isolate the parameter $m$, perform reasonable substitution, and determine the range of $m$ through extremum discussion.
Solution: The problem is equivalent to
$\cos 2 \theta-3>2 m \cos \theta-4 m$.
Since $2-\cos \theta \in[1,3]$, then
$$
\begin{array}{l}
m>\frac{2-\cos ^{2} \theta}{2-\cos \theta}=2+\cos \theta-\frac{2}{2-\cos \theta} \\
=4-\left((2-\cos \theta)+\frac{2}{2-\cos \theta}\right]
\end{array}
$$

always holds.
Let $t=2-\cos \theta$, then $t \in[1,3]$. The original problem is equivalent to $t+\frac{2}{t}>4-m$ always holds.
Since when $t \in[1,3]$, $t+\frac{2}{t} \geqslant 2 \sqrt{2}$,
Therefore, for the original problem to always hold, $m>4-2 \sqrt{2}$.