Example 12 Given $f(\theta)=\sin ^{2} \theta+\sin ^{2}(\theta+\alpha)$ $+\sin ^{2}(\theta+\beta)$, where $\alpha, \beta$ are constants satisfying $0 \leqslant \alpha \leqslant \beta \leqslant \pi$. For what values of $\alpha, \beta$ is $f(\theta)$ a constant value?
To make the value of $f(\theta)$ not change with $\theta$, i.e., the function $f(\theta)$ is a constant function, we can assign special values to the independent variable to explore.
$$
\begin{array}{l}
\text { Solution: Let } \theta=0, \frac{\pi}{6}, \frac{\pi}{2} \text {, we get } \\
f(0)=\sin ^{2} \alpha+\sin ^{2} \beta, \\
f\left(\frac{\pi}{6}\right)=\frac{1}{4}+\sin ^{2}\left(\frac{\pi}{6}+\alpha\right)+\sin ^{2}\left(\frac{\pi}{6}+\beta\right), \\
f\left(\frac{\pi}{2}\right)=1+\cos ^{2} \alpha+\cos ^{2} \rho .
\end{array}
$$

From the problem, we can assume
$f(0)=f\left(\frac{\pi}{6}\right)=f\left(\frac{\pi}{2}\right)=m (m$ is a constant),
then from $f(0)+f\left(\frac{\pi}{2}\right)=2 m$ we get $m=\frac{3}{2}$.
Substituting $f(0)=f\left(\frac{\pi}{6}\right)=f\left(\frac{\pi}{2}\right)=\frac{3}{2}$, we solve to get $\alpha=\frac{\pi}{3}, \beta=\frac{2 \pi}{3}$.