Quay lại đáp án

$1+\sin 2x \sin x -\cos x \sin^2 2x = 1+\cos (\dfrac{\pi}{2}-2x)=1+\sin 2x$$\Leftrightarrow \sin 2x \sin x -\sin^2 2x \cos x -\sin 2x =0$$\Leftrightarrow \sin 2x (\sin x -\sin 2x \cos x-1)=0$$\Leftrightarrow \sin 2x(\sin x -1 -2\sin^2 x \cos x)=0$$\Leftrightarrow \sin 2x [\sin x (1-2\sin^2 x)-1]=0 $$\Leftrightarrow \sin 2x ( \sin x \cos 2x -1)=0$TH1 $\sin 2x =0$ dễTH2 $ \sin x \cos 2x =1$$ \Leftrightarrow \sin 3x -\sin x = 2 \Leftrightarrow $$\begin{cases}\sin 3x = 1 \\ \sin x =-1 \end{cases}$$ $vô nghiệmKL pt có nghiệm$\sin 2x = 0 \Leftrightarrow x =\dfrac{k\pi}{2};\ k \in Z$

$1+\sin 2x \sin x -\cos x \sin^2 2x = 1+\cos (\dfrac{\pi}{2}-2x)=1+\sin 2x$$\Leftrightarrow \sin 2x \sin x -\sin^2 2x \cos x -\sin 2x =0$$\Leftrightarrow \sin 2x (\sin x -\sin 2x \cos x-1)=0$$\Leftrightarrow \sin 2x(\sin x -1 -2\sin x \cos ^2x)=0$$\Leftrightarrow \sin 2x [\sin x (1-2\cos^2 x)-1]=0$$\Leftrightarrow \sin 2x ( -\sin x \cos 2x -1)=0$TH1$\sin 2x =0$dễTH2$ \sin x \cos 2x =-1$$\Leftrightarrow \sin 3x -\sin x = -2 \Leftrightarrow \begin{cases}\sin 3x = -1 \\ \sin x =-1 \end{cases}$vô nghiệmKL pt có nghiệm$\sin 2x = 0 \Leftrightarrow x =\dfrac{k\pi}{2};\ k \in Z$