4. In $\triangle A B C$, it is known that
$$
\begin{array}{l}
(\sqrt{3} \sin B-\cos B)(\sqrt{3} \sin C-\cos C) \\
=4 \cos B \cdot \cos C,
\end{array}
$$

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. C.

From the given, we have
$$
\begin{array}{l}
\sqrt{3}(\sin B \cdot \cos C+\cos B \cdot \sin C) \\
=-3(\cos B \cdot \cos C-\sin A \cdot \sin C) \\
\Rightarrow \sqrt{3} \sin (B+C)=-3 \cos (B+C) \\
\Rightarrow \tan (B+C)=-\sqrt{3} .
\end{array}
$$

Since $0<\angle B+\angle C<\pi$, we have
$$
\angle B+\angle C=\frac{2 \pi}{3}, \angle A=\frac{\pi}{3} \text {. }
$$

From $A B+A C=4$, we get $A B=4-A C$. Therefore,
$$
\begin{array}{l}
B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cos A \\
=(4-A C)^{2}+A C^{2}-(4-A C) A C \\
=3(A C-2)^{2}+4 \geqslant 4 .
\end{array}
$$

The equality holds if and only if $A C=2$.
Thus, $B C \geqslant 2$.
Also, $B C<A B+A C=4$, so the range of $B C$ is $[2,4)$.