Example 4 Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x<0$, $f(x)$ is monotonically increasing, $f(-1)=0$. Let
$$
\begin{array}{l}
\varphi(x)=\sin ^{2} x+m \cos x-2 m, \\
M=\left\{m \mid \text { for any } x \in\left[0, \frac{\pi}{2}\right], \varphi(x)<0\right\}, \\
N=\left\{m \mid \text { for any } x \in\left[0, \frac{\pi}{2}\right], f(\varphi(x))<0\right\} .
\end{array}
$$

Find $M \cap N$.
Notice that,
$$
\begin{array}{l}
\varphi(x)=-\cos ^{2} x+m \cos x+(1-2 m) \\
=-t^{2}+m t+(1-2 m)=g(t),
\end{array}
$$

where $t=\cos x \in[0,1]$.
Then for any $x \in\left[0, \frac{\pi}{2}\right]$, we have
$\varphi(x)\max _{t \in[0,1]} \frac{1-t^{2}}{2-t}$.
And $\frac{1-t^{2}}{2-t}=4-\left(2-t+\frac{3}{2-t}\right) \leqslant 4-2 \sqrt{3}$.
Therefore, $M=[4-2 \sqrt{3},+\infty)$.
By the problem's condition, using the graph it is easy to see that
$f(x)0$, which is a contradiction.
If $00$ or $g\left(\frac{m}{2}\right)0$, which is a contradiction.
Hence, $g\left(\frac{m}{2}\right)=\frac{m^{2}}{4}-2 m+11$, then $g(0)>0$ or $g(1)0 \Rightarrow m1 \Rightarrow m>2$.
In summary, $N=(4-2 \sqrt{2},+\infty)$.
Thus, $M \cap N=(4-2 \sqrt{2},+\infty)$.