已知 $-\dfrac{\pi}{2}<\beta-\alpha<\dfrac{\pi}{2}$，$\sin \beta-2 \cos \alpha=1$，$2 \sin \alpha+\cos \beta=\sqrt{2}$，则 $\sin \left(\alpha+\dfrac{\pi}{6}\right)=$ （ ）
A．$\dfrac{\sqrt{3}}{3}$
B．$\dfrac{\sqrt{6}}{3}$
C．$\pm \dfrac{\sqrt{6}}{3}$
D．$\pm \dfrac{\sqrt{3}}{3}$
答案 B．
解析 根据题意，有\[\begin{cases} \sin\beta=2\cos\alpha-1,\\ \cos\beta=\sqrt 2-2\sin\alpha,\end{cases}\implies (2\cos\alpha-1)^2+(\sqrt 2-2\sin\alpha)^2=1,\]其中 $\cos\alpha\in (0,1)$ 且 $\sin\alpha\in \left(\dfrac{\sqrt 2-1}2,\dfrac{\sqrt 2+1}2\right)$，即\[4\sqrt 2\sin\alpha+4\cos\alpha=6\iff \sin\left(\alpha+\arctan\dfrac{\sqrt 2}2\right)=\dfrac{\sqrt 3}2\iff \alpha=\dfrac{\pi}3-\arctan\dfrac{\sqrt 2}2,\]因此\[\sin\left(\alpha+\dfrac{\pi}6\right)=\sin\left(\dfrac{\pi}2-\arctan\dfrac{\sqrt 2}2\right)=\dfrac{\sqrt 6}3.\]