Example 2 In $\triangle A B C$, it is known that $a=2 b \cos C, \sin A \sin \left(\frac{B}{2}+C\right)=\sin C\left(\sin \frac{B}{2}+\sin A\right)$, find the degree measures of the interior angles of this triangle. (Do not use inverse trigonometric functions to represent).
Let the circumradius of $\triangle ABC$ be $R$. According to the given $a=2b\cos C$, and the projection theorem $a=b\cos C+c\cos B$, we know that $b\cos C=c\cdot\cos B$, which means $\sin B\cos C=\sin C\cos B$. Thus, $\sin (B-C)=0$, and since $B-C \in (-\pi, \pi)$, we have $B=C, A=\pi-2B$. Substituting into $\sin A \sin \left(\frac{B}{2}+C\right)=\sin C\left(\sin \frac{B}{2}+\sin A\right)$, we get $\sin \frac{B}{4} \cos \frac{9B}{4}=0$. Since $0<B<\frac{\pi}{2}$, then $\cos \frac{9B}{4}=0$, hence $B=\frac{2\pi}{9}$, and thus, $A=\frac{5}{9}\pi, B=C=\frac{2\pi}{9}$.