Giải phương trình 4sinx.cos(x-pi/2)+4sin(pi+x).cosx+2sin(3pi/2-x).cos(pi+x)=1

$4\sin x.\sin x -4\sin x\cos x+2\sin(\dfrac{-\pi}{2}-x).(-\cos x)=1$

$\Leftrightarrow 4\sin^2x-4\sin x\cos x+2\cos^2x=1$    (*)

ĐK: $x\ne \dfrac{\pi}{2}+k\pi$

(*) $\Leftrightarrow 4\tan^2x-4\tan x+2=1+\tan^2x$

$\Leftrightarrow 3\tan^2x-4\tan x+1=0$

$\Leftrightarrow \tan x=1$ hoặc $\tan x=\dfrac{1}{3}$

$\Leftrightarrow x=\dfrac{\pi}{4}+k\pi$ hoặc $x=\arctan\dfrac{1}{3}+k\pi$ (TM)