sin22x−2sin22x−4cos2x=tan2xĐK $\sin^2 2x -4\cos^2 x \ne 0$$\Leftrightarrow 4\cos^2 x(\sin^2 x - 1) \ne 0$$\Leftrightarrow -4\cos^4 x \ne 0$$\Leftrightarrow \cos x \ne 0 \Leftrightarrow x\ne \dfrac{\pi}{2}+k\pi;\ \ k \in Z$Vậy pt đã cho tương đương $\sin^2 2x -2 = \tan^2 x.(-4\cos^4 x) = -4\sin^2 x \cos^2 x=-\sin^2 2x$$\Leftrightarrow 2\sin^2 2x - 2 = 0$$\Leftrightarrow \sin^2 2x - 1 = 0$$\Leftrightarrow \cos^2 2x = 0 \Leftrightarrow \cos 2x = 0$$\Leftrightarrow x=\dfrac{\pi}{4} +\dfrac{k\pi}{2};\ k\in Z$
sin22x−2sin22x−4cos2x=tan2xĐK $\sin^2 2x -4\cos^2 x \ne 0$$\Leftrightarrow 4\cos^2 x(\sin^2 x - 1) \ne 0$$\Leftrightarrow -4\cos^4 x \ne 0$$\Leftrightarrow \cos x \ne 0 \Leftrightarrow x\ne \dfrac{\pi}{2}+k\pi;\ \ k \in Z$Vậy pt đã cho tương đương $\sin^2 2x -2 = \tan^2 x.(-4\cos^3 x) = -4\sin^2 x \cos^2 x=-\sin^2 2x$$\Leftrightarrow 2\sin^2 2x - 2 = 0$$\Leftrightarrow \sin^2 2x - 1 = 0$$\Leftrightarrow \cos^2 2x = 0 \Leftrightarrow \cos 2x = 0$$\Leftrightarrow x=\dfrac{\pi}{4} +\dfrac{k\pi}{2};\ k\in Z$
sin22x−2sin22x−4cos2x=tan2xĐK $\sin^2 2x -4\cos^2 x \ne 0$$\Leftrightarrow 4\cos^2 x(\sin^2 x - 1) \ne 0$$\Leftrightarrow -4\cos^4 x \ne 0$$\Leftrightarrow \cos x \ne 0 \Leftrightarrow x\ne \dfrac{\pi}{2}+k\pi;\ \ k \in Z$Vậy pt đã cho tương đương $\sin^2 2x -2 = \tan^2 x.(-4\cos^
4 x) = -4\sin^2 x \cos^2 x=-\sin^2 2x$$\Leftrightarrow 2\sin^2 2x - 2 = 0$$\Leftrightarrow \sin^2 2x - 1 = 0$$\Leftrightarrow \cos^2 2x = 0 \Leftrightarrow \cos 2x = 0$$\Leftrightarrow x=\dfrac{\pi}{4} +\dfrac{k\pi}{2};\ k\in Z$