Example 2 (2003 National College Entrance Examination - Jiangsu Paper) Given the function $f(x)=\sin (\omega x+\varphi)(\omega>0,0 \leqslant \varphi \leqslant \pi)$ is an even function on $\mathbf{R}$, its graph is symmetric about the point $M\left(\frac{3}{4} \pi, 0\right)$, and it is a monotonic function in the interval $\left[0, \frac{\pi}{2}\right]$, find the values of $\omega$ and $\varphi$.
Analysis: This problem should make full use of its symmetry, and continuously observe and explore while solving it.
Solution: Given that $f(x)$ is an even function and considering the properties of sine and cosine trigonometric functions, we have
$f(0)= \pm 1$ which means $\sin \varphi= \pm 1$. Since $0 \leqslant \varphi \leqslant \pi$, we have $\varphi=\frac{\pi}{2}$.
Also, the graph of $f(x)$ is symmetric about $\left(\frac{3}{4} \pi, 0\right)$ and from the properties of the function graph: $f\left(\frac{3 \pi}{4}\right)=0$ which means $\sin \left(\frac{3}{4} \pi \omega+\frac{\pi}{2}\right)=0$.
Thus, $\frac{3}{4} \pi \omega+\frac{\pi}{2}=k \pi$ so $\omega=\frac{2}{3}(2 k-1)$.
Since $\omega>0$, we have $k=1,2,3, \cdots$.
When $k=1$, $\omega=\frac{2}{3}$, and $f(x)=\sin \left(\frac{2}{3} x+\frac{\pi}{2}\right)$ is a decreasing function on $\left[0, \frac{\pi}{2}\right]$.
When $k=2$, $\omega=2$, and $f(x)=\sin \left(2 x+\frac{\pi}{2}\right)$ is a decreasing function on $\left[0, \frac{\pi}{2}\right]$.
When $k \geqslant 3$, $\omega \geqslant \frac{10}{3}$, and $f(x)=\sin \left(\omega x+\frac{\pi}{2}\right)$ is not a monotonic function on $\left[0, \frac{\pi}{2}\right]$.
In summary, $\omega=\frac{2}{3}$ or $\omega=2$, and $\varphi=\frac{\pi}{2}$.