Giải phương trình: sin2x - (sinx + cosx - 1)(2sinx - cosx - 3) = 0

Đáp án: $x=\dfrac12k\pi$

Giải thích các bước giải:

Ta có:

$\sin2x-(\sin x+\cos x-1)(2\sin x-\cos x-3)=0$

$\to \sin2x=(\sin x+\cos x-1)(2\sin x-\cos x-3)$

$\to 2\sin x\cos x=(\sin x+\cos x-1)(2\sin x-\cos x-3)$

$\to (2\sin x\cos x+1)-1=(\sin x+\cos x-1)(2\sin x-\cos x-3)$

$\to (2\sin x\cos x+\sin^2x+\cos^2x)-1=(\sin x+\cos x-1)(2\sin x-\cos x-3)$

$\to (\sin x+\cos x)^2-1=(\sin x+\cos x-1)(2\sin x-\cos x-3)$

$\to (\sin x+\cos x-1)(\sin x+\cos x+1)=(\sin x+\cos x-1)(2\sin x-\cos x-3)$

$\to (\sin x+\cos x-1)(\sin x+\cos x+1)-(\sin x+\cos x-1)(2\sin x-\cos x-3)=0$

$\to (\sin x+\cos x-1)(\sin x+\cos x+1-2\sin x+\cos x+3)=0$

$\to (\sin x+\cos x-1)(2\cos x-\sin x+4)=0$

Mà $2\cos x-\sin x+4\ge -2-1+4>0$

$\to \sin x+\cos x-1=0$

$\to \sin x+\cos x=1$

$\to (\sin x+\cos x)^2=1$

$\to 1+\sin2x=1$

$\to \sin2x=0$

$\to 2x=k\pi$

$\to x=\dfrac12k\pi$