Đặt $x=t+\frac{\pi}{6}$
pt đã cho $\Leftrightarrow 4\sin (t+\frac{\pi}{3}).(\sin (2t+\frac{\pi}{2})-1)=2\cos (2t+\frac{\pi}{3})-1$
$\Leftrightarrow 4\sin (t+\frac{\pi}{3})(\cos 2t-1)=2(\cos 2t.\frac{1}{2}-\sin 2t.\frac{\sqrt{3}}{2})-1$
$\Leftrightarrow 4\sin (t+\frac{\pi}{3})(\cos2t-1)=(2\cos 2t-1)-\sqrt{3}\sin 2t$
$\Leftrightarrow [4\sin (t+\frac{\pi}{3})-1](\cos 2t-1)=-\sqrt{3}\sin 2t$
$\Leftrightarrow [4\sin (t+\frac{\pi}{3})-1](-2\sin ^2t)=-2\sqrt{3}\sin t\cos t$
$\Leftrightarrow [\begin{matrix} \sin t=0 (1)\\ -2\sin t(2\sin t+2\sqrt{3}\cos t-1)=-2\sqrt{3}\cos t (2)\end{matrix}$
$(2)\Leftrightarrow -2\sin t(2\sin t-1)=-2\sqrt{3}\cos t+4\sqrt{3}\sin t\cos t$
$\Leftrightarrow -2\sin t(2\sin t-1)=2\sqrt{3}\cos t(2\sin t-1)$
$\Leftrightarrow [\begin{matrix} 2\sin t=1 (3)\\ -\sin t=\sqrt{3}\cos t (4)\end{matrix}$
Giải (1)(3)(4) tìm được nghiệm