giải pt 20/ $\frac{1+sin2x+cos2x}{1+cot^2x}= √2sinx.sin2x$

ĐK: $\sin x\ne 0\Leftrightarrow x\ne k\pi$

$\dfrac{1+\sin2x+2\cos^2x-1}{\dfrac{1}{\sin^2x}}$

$=\sin^2x(2\cos^2x+\sin2x)$

$\Rightarrow \sin^2x(2\cos^2x+\sin2x)-\sqrt2\sin x.\sin2x=0$

$\Leftrightarrow \sin^2x(2\cos^2x+\sin2x-2\sqrt2\cos x)=0$

$\Leftrightarrow 2\sin^2x(\cos^2x+\sin x\cos x+\sqrt2\cos x)=0$

$\Leftrightarrow 2\sin^2x\cos x(\cos x+\sin x+\sqrt2)=0$

$+) \sin x=0\Leftrightarrow x=k\pi$ (loại)

$+) \cos x=0\Leftrightarrow x=\dfrac{\pi}{2}+k\pi$ (TM)

$+) \cos x+\sin x=-\sqrt2\Leftrightarrow \sqrt2\sin(x+\dfrac{\pi}{4})=-\sqrt2\Leftrightarrow \sin(x+\dfrac{\pi}{4})=-1\Leftrightarrow x=\dfrac{-3\pi}{4}+k2\pi$ (TM)