$\sin x\cos x = \dfrac{\sqrt{3}}{4} \Rightarrow \dfrac{\sin 2x}{2} = \dfrac{\sqrt{3}}{4} \Rightarrow \sin 2x = \dfrac{\sqrt{3}}{2}$. So $2x = \dfrac{\pi}{3}+2n\pi$ or $2x = \pi-\dfrac{\pi}{3}+2n\pi = \dfrac{2\pi}{3}+2n\pi$. Thus $x = \dfrac{\pi}{6}+n\pi$ or $x = \dfrac{\pi}{3}+n\pi$... Hmm, or $x = \dfrac{\pi}{12}+n\pi$ or $x = \dfrac{5\pi}{12}+n\pi$. Dividing by 2: $x = \dfrac{\pi}{6}+n\pi$ or $x=\dfrac{\pi}{3}+n\pi$. Per key: option (A).