Chọn phương án B.

\(\begin{aligned}
\sin\left(2x-40^\circ\right)=\dfrac{\sqrt{3}}{2}\Leftrightarrow&\left[\begin{array}{l}2x-40^\circ=60^\circ+k360^\circ\\ 2x-40^\circ=120^\circ+k360^\circ\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}2x=100^\circ+k360^\circ\\ 2x=160^\circ+k360^\circ\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=50^\circ+k180^\circ\\ x=80^\circ+k180^\circ\end{array}\right.
\end{aligned}\)

\(\blacksquare\) Với \(x=50^\circ+k180^\circ\): $$\begin{eqnarray*}
-180^\circ\leq&x&\leq180^\circ\\
\Leftrightarrow-180^\circ\leq&50^\circ+k180^\circ&\leq180^\circ\\
\Leftrightarrow-230^\circ\leq&k180^\circ&\leq130^\circ\\
\Leftrightarrow-\dfrac{23}{18}\leq&k&\leq\dfrac{13}{18}
\end{eqnarray*}$$Vì \(k\in\mathbb{Z}\) nên \(\left[\begin{array}{l}k=-1\\ k=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-130^\circ\\ x=50^\circ\end{array}\right.\)

\(\blacksquare\) Với \(x=80^\circ+k180^\circ\): $$\begin{eqnarray*}
-180^\circ\leq&x&\leq180^\circ\\
\Leftrightarrow-180^\circ\leq&80^\circ+k180^\circ&\leq180^\circ\\
\Leftrightarrow-260^\circ\leq&k180^\circ&\leq100^\circ\\
\Leftrightarrow-\dfrac{13}{9}\leq&k&\leq\dfrac{5}{9}
\end{eqnarray*}$$Vì \(k\in\mathbb{Z}\) nên \(\left[\begin{array}{l}k=-1\\ k=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-100^\circ\\ x=80^\circ\end{array}\right.\)

Vậy phương trình đã cho có \(4\) nghiệm trên đoạn \(\left[-180^\circ;180^\circ\right]\).