Chọn phương án C.

\(\begin{aligned}
\sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\Leftrightarrow&\left[\begin{array}{l}2x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\ 2x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}2x=-\dfrac{\pi}{6}+k2\pi\\ 2x=\dfrac{\pi}{2}+k2\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{ll}x=-\dfrac{\pi}{12}+k\pi &(1)\\ x=\dfrac{\pi}{4}+k\pi &(2)\end{array}\right.
\end{aligned}\)

• Vì \(x=-\dfrac{\pi}{12}+k\pi=-\dfrac{\pi}{12}+k\dfrac{2\pi}{2}\) nên (1) có \(2\) điểm biểu diễn.
• Vì \(x=\dfrac{\pi}{4}+k\pi=\dfrac{\pi}{4}+k\dfrac{2\pi}{2}\) nên (2) có \(2\) điểm biểu diễn.