(4) In $\triangle A B C$,
$$
(\sqrt{3} \sin B-\cos B)(\sqrt{3} \sin C-\cos C)=4 \cos B \cos C,
$$

and $A B+A C=4$, then the range of $B C$ is ( ).
(A) $(2,4)$
(B) $(2,4]$
(C) $[2,4)$
(D) $[2,4]$
(4) $\mathrm{C}$ Hint: Method 1: From the given
$$
\begin{aligned}
& 3 \sin B \sin C-\sqrt{3} \sin B \cos C-\sqrt{3} \cos B \sin C+\cos B \cos C \\
= & 4 \cos B \cos C,
\end{aligned}
$$
- $\sqrt{3}(\sin B \cos C+\cos B \sin C)=-3(\cos B \cos C-\sin A \sin C)$,
$$
\begin{aligned}
\sqrt{3} \sin (B+C) & =-3 \cos (B+C) \\
\tan (B+C) & =-\sqrt{3}
\end{aligned}
$$

Since $0<B+C<\pi$, we have $B+C=\frac{2 \pi}{3}, A=\pi-(B+C)=\frac{\pi}{3}$.
From $A B+A C=4$, we get $A B=4-A C$.
$$
\begin{aligned}
B C^{2} & =A B^{2}+A C^{2}-2 A B \cdot A C \cos A \\
& =(4-A C)^{2}+A C^{2}-(4-A C) \cdot A C \\
& =3 A C^{2}-12 A C+16 \\
& =3(A C-2)^{2}+4 \\
& \geqslant 4 .
\end{aligned}
$$

When and only when $A C=2$, the above inequality holds with equality, so $B C \geqslant 2$. Also, $B C<A B+A C=4$. Therefore, the range of $B C$ is $[2,4)$. Hence, the answer is $C$.
Method 2: From the given
$$
\begin{array}{l}
(\sqrt{3} \tan B-1)(\sqrt{3} \tan C-1)=4, \\
3 \tan B \tan C-\sqrt{3} \tan B-\sqrt{3} \tan C+1=4, \\
\sqrt{3}(\tan B+\tan C)=3(\tan B \tan C-1), \\
\tan (B+C)=\frac{\tan B+\tan C}{1-\tan B \tan C}=-\sqrt{3} .
\end{array}
$$

The rest is the same as Method 1.