Example 4 In $\triangle A B C$, it is known that
$$
\begin{array}{l}
\sin \left(\frac{B}{2}+C\right)|\sin A-\sin C|=\sin \frac{B}{2} \cdot \sin A, \\
a=2 b \cos C .
\end{array}
$$

Find the three interior angles of this triangle.
Triangle Solution From the given condition $a=2 b \cos C$, and the projection theorem, we have
$$
\begin{array}{l}
a=b \cos C+c \cos B=2 b \cos C \\
\Leftrightarrow b \cos C=c \cos B .
\end{array}
$$

By the sine rule, we get
$$
\sin B \cdot \cos C-\sin C \cdot \cos B=0 \Rightarrow \sin (B-C)=0 \text {. }
$$

Therefore, $\angle B=\angle C, \angle A=\pi-2 \angle B$.
Substituting the above conclusion into the given equation, we get
$$
\begin{array}{l}
\sin \frac{3 B}{2}|\sin 2 B-\sin B|=\sin 2 B \cdot \sin \frac{B}{2} \\
\Leftrightarrow\left|\sin 2 B \cdot \sin \frac{3 B}{2}-\sin B \cdot \sin \frac{3 B}{2}\right|=\sin 2 B \cdot \sin \frac{B}{2} \\
\Leftrightarrow\left|\cos \frac{B}{2}-\cos \frac{7 B}{2}+\cos \frac{5 B}{2}-\cos \frac{B}{2}\right| \\
\quad=\cos \frac{3 B}{2}-\cos \frac{5 B}{2} \\
\Leftrightarrow\left|\cos \frac{5 B}{2}-\cos \frac{7 B}{2}\right|=\cos \frac{3 B}{2}-\cos \frac{5 B}{2} \\
\Leftrightarrow \cos \frac{5 B}{2}-\cos \frac{7 B}{2}=\cos \frac{3 B}{2}-\cos \frac{5 B}{2} \text { or } \\
\quad \cos \frac{7 B}{2}-\cos \frac{5 B}{2}=\cos \frac{3 B}{2}-\cos \frac{5 B}{2} \\
\Leftrightarrow \cos \frac{3 B}{2}+\cos \frac{7 B}{2}-2 \cos \frac{5 B}{2}=0 \text { or } \\
\cos \frac{7 B}{2}-\cos \frac{3 B}{2}=0 .
\end{array}
$$
$\Leftrightarrow \cos \frac{3 B}{2}+\cos \frac{7 B}{2}-2 \cos \frac{5 B}{2}=0$ or

Given $0<60^{\circ}$,
$B C>A B$.

Taking a point $E$ on side $B C$ such that $C D=C E$. Then
$$
\begin{array}{l}
B A=B E, \triangle A B D \cong \triangle E B D . \\
\text { Hence } \angle A=\angle B E D \\
=180^{\circ}-\left(90^{\circ}-\frac{\angle C}{2}\right) \\
=90^{\circ}+\frac{\angle C}{2} .
\end{array}
$$

Also, $\angle A=180^{\circ}-2 \angle C$, thus,
$180^{\circ}-2 \angle C=90^{\circ}+\frac{\angle C}{2}$
$$
\Rightarrow \angle C=36^{\circ} \Rightarrow \angle A=108^{\circ} \text {. }
$$
(2) If $A B>B C$, then take a point $D$ on $A C$ such that $C D=A B-B C$, as shown in Figure 5.
Clearly, $\angle A<60^{\circ}, B C<A B$.
Taking a point $E$ on side $A B$ such that $B C=$ $B E$, and connecting $D E$.
Hence $\triangle B D C \cong \triangle B D E$
$$
\begin{array}{l}
\Rightarrow \angle C=\angle B E D . \\
\text { Also, } C D=E D=A E, \\
\angle B E D=\angle A+\angle A D E,
\end{array}
$$

Therefore, $\angle A=\angle A D E, \angle C=2 \angle A$.
Thus, $5 \angle A=180^{\circ} \Rightarrow \angle A=36^{\circ}$
$$
\Rightarrow \angle B=\angle C=72^{\circ}
$$