Chọn phương án A.

$\begin{aligned}
\sin x-\sqrt{3}\cos x=2\Leftrightarrow&\dfrac{1}{2}\sin x-\dfrac{\sqrt{3}}{2}\cos x=1\\
\Leftrightarrow&\sin\left(x-\dfrac{\pi}{3}\right)=1\\ \Leftrightarrow&x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\\ \Leftrightarrow&x=\dfrac{5\pi}{6}+k2\pi\;(k\in\mathbb{Z}).
\end{aligned}$

Trên khoảng $(0;5\pi)$ ta có $$\begin{aligned}
0<\dfrac{5\pi}{6}+k2\pi<5\pi\Leftrightarrow& -\dfrac{5}{12}< k<\dfrac{25}{12}\;(k\in\Bbb{Z})\\
\Leftrightarrow&\left[\begin{array}{l}
k=0\\ k=1\\ k=2
\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}
x=\dfrac{5\pi}{6}\\ x=\dfrac{17\pi}{6}\\ x=\dfrac{29\pi}{6}
\end{array}\right.
\end{aligned}$$
Vậy phương trình có $3$ nghiệm thuộc khoảng $(0;5\pi)$.