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Câu 1:$PT\Leftrightarrow \frac{2sin x}{sin 2x}+\frac{1}{sin 2x}=\frac{1}{sin 2x.cos 2x}$ (sin 2x kahcs 0,cos 2x khác 0)$\Leftrightarrow (2sin x+1).cos 2x=1$$\Leftrightarrow 2sin x.cos 2x+cos 2x=sin^2 x+cos^2 x$$\Leftrightarrow sin x(2cos 2x-sin x)+cos^2 x-1=0$$\Leftrightarrow sin x(2cos 2x-sin x)-sin^2 x=0$$\Leftrightarrow sin x(2cos 2x-2sin x)=0$$\Leftrightarrow sin x(-4sin^2 x-2sin x+2)=0$

Câu 1:$PT\Leftrightarrow \frac{2sin x}{sin 2x}+\frac{1}{sin 2x}=\frac{1}{sin 2x.cos 2x}$ (sin 2x khác 0,cos 2x khác 0)$\Leftrightarrow (2sin x+1).cos 2x=1$$\Leftrightarrow 2sin x.cos 2x+cos 2x=sin^2 x+cos^2 x$$\Leftrightarrow sin x(2cos 2x-sin x)+cos^2 x-1=0$$\Leftrightarrow sin x(2cos 2x-sin x)-sin^2 x=0$$\Leftrightarrow sin x(2cos 2x-2sin x)=0$$\Leftrightarrow sin x(-4sin^2 x-2sin x+2)=0$