a) ĐK: $\sin x\neq \frac{1}{2}$PT $\Leftrightarrow \frac{\tan x(4\cos^3 x-3\cos x)+2(2\cos^2 x-1)-1}{1-2\sin x}=\sqrt{3}\cos x(2\sin x+1)$
$\Leftrightarrow (4\cos^2x\sin x+4\cos^2 x)-(3\sin x+3)=\sqrt{3}\cos x(1-4\sin^2 x)$
$\Leftrightarrow (\sin x+1)(4\cos ^2x-3)=\sqrt{3}\cos x(4\cos ^2x-3)$
$\Leftrightarrow (4\cos^2 x-3)(\sin x-\sqrt{3}\cos x+1)=0$
$\Leftrightarrow 4\cos ^2x-3=0....or...\sin x-\sqrt{3}\cos x+1=0$( Giải PT này = cách chia cả PT cho 2 )
Tự làm tiếp nhé !!!