Đáp án: $x=\dfrac12k\pi$
Giải thích các bước giải:
Ta có:
$|\sin x-\cos x|+4\sin2x=1$
$\to |\sin x-\cos x|+8\sin x\cos x=1$
$\to |\sin x-\cos x|+8\sin x\cos x-4=-3$
$\to |\sin x-\cos x|-4(1-2\sin x\cos x)=-3$
$\to |\sin x-\cos x|-4(\sin^2x+\cos^2x-2\sin x\cos x)=-3$
$\to |\sin x-\cos x|-4(\sin x-\cos x)^2=-3$
$\to 4(\sin x-\cos x)^2-|\sin x-\cos x|-3=0$
$\to 4|\sin x-\cos x|^2-|\sin x-\cos x|-3=0$
$\to (|\sin x-\cos x|-1)(4|\sin x-\cos x|+3)=0$
$\to |\sin x-\cos x|-1=0$ vì $4|\sin x-\cos x|+3\ge 0+3>0$
$\to |\sin x-\cos x|=1$
$\to |\sin x-\cos x|^2=1$
$\to (\sin x-\cos x)^2=1$
$\to \sin^2x+\cos^2x-2\sin x\cos x=1$
$\to 1-\sin2x=1$
$\to \sin2x=0$
$\to 2x=k\pi$
$\to x=\dfrac12k\pi$
