Giải thích các bước giải:
Ta có:
$\cos2x\sin2x=\dfrac{\sqrt{3}}{4}$
$\to 2\cos2x\sin2x=\dfrac{\sqrt{3}}{2}$
$\to \sin4x=\dfrac{\sqrt{3}}{2}$
$\to 4x=\dfrac13\pi+k2\pi\to x=\dfrac1{12}\pi+\dfrac12k\pi$
Hoặc $4x=\dfrac23\pi+k2\pi\to x=\dfrac16\pi+\dfrac12k\pi$