Example 4 Given that $\alpha$ and $\beta$ are acute angles, and $\cos \alpha + \cos \beta - \cos (\alpha + \beta) = \frac{3}{2}$, find the values of $\alpha$ and $\beta$.


Solution 1: From the problem, we have
$4 \cos ^{2} \frac{\alpha+\beta}{2}-4 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}+1=0$, which means
( $\left.2 \cos \frac{\alpha+\beta}{2}-\cos \frac{\alpha-\beta}{2}\right)^{2}+\sin ^{2} \frac{\alpha-\beta}{2}=0$.
Thus, $2 \cos \frac{\alpha+\beta}{2}-\cos \frac{\alpha-\beta}{2}=0$, and $\sin \frac{\alpha-\beta}{2}=0$.
Since $\alpha, \beta$ are acute angles, i.e., $-\frac{\pi}{2}<\alpha-\beta<\frac{\pi}{2}$, then $\alpha=\beta$, leading to $\alpha=\beta$ $=\frac{\pi}{3}$.
Solution 2: From the problem, we have
$$
\sin \alpha \sin \beta+(1-\cos \alpha) \cos \beta+\cos \alpha-\frac{3}{2}=0 \text {. }
$$

From $\sin ^{2} \beta+\cos ^{2} \beta=1$, we know that the point $(\cos \beta, \sin \beta)$ is the intersection of the line $(1-\cos \alpha) x+$ $y \sin \alpha+\cos \alpha-\frac{3}{2}=0$ and the circle $x^{2}+y^{2}=1$. Therefore, the distance from the center $(0,0)$ to this line is no greater than the radius of the circle, i.e., $d=\frac{\left|\cos \alpha-\frac{3}{2}\right|}{\sqrt{\sin ^{2} \alpha+(1-\cos \alpha)^{2}}} \leqslant 1$, simplifying to $\left(\cos \alpha-\frac{1}{2}\right)^{2} \leqslant 0$.
Thus, $\cos \alpha=\frac{1}{2}$, yielding $\alpha=\frac{\pi}{3}$. Similarly, $\beta=\frac{\pi}{3}$.