Chọn phương án B.

\begin{eqnarray*}
\sqrt{2}\cos\left(x+\dfrac{\pi}{3}\right)=1&\Leftrightarrow & \cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{2}}{2}\\
&\Leftrightarrow&\cos\left(x+\dfrac{\pi}{3}\right)=\cos \dfrac{\pi}{4} \\
&\Leftrightarrow &\left[\begin{array}{l}
x+\dfrac{\pi}{3}=\dfrac{\pi}{4}+k2\pi\\ x+\dfrac{\pi}{3}=-\dfrac{\pi}{4}+k2\pi
\end{array}\right.\\
&\Leftrightarrow&\left[\begin{array}{l}
x=-\dfrac{\pi}{12}+k2\pi\\ x=-\dfrac{7\pi}{12}+k2\pi
\end{array}\right.\;(k\in\mathbb{Z}).
\end{eqnarray*}

• Với $x=-\dfrac{\pi}{12}+k2\pi$: $$0\leq-\dfrac{\pi}{12}+k2\pi\leq2\pi\Leftrightarrow\dfrac{1}{24}\leq k\leq\dfrac{25}{24}$$Vì $k\in\mathbb{Z}$ nên $k=1\Rightarrow x=\dfrac{23\pi}{12}$.
• Với $x=-\dfrac{7\pi}{12}+k2\pi$: $$0\leq-\dfrac{7\pi}{12}+k2\pi\leq2\pi\Leftrightarrow\dfrac{7}{24}\leq k\leq\dfrac{31}{24}$$Vì $k\in\mathbb{Z}$ nên $k=1\Rightarrow x=\dfrac{17\pi}{12}$.

Vậy phương trình có $2$ nghiệm trên đoạn $[0;2\pi]$.