Điều kiện \sin 2x \ne 0 \Leftrightarrow x\ne \dfrac{k\pi}{2};\ k\in ZPT \Leftrightarrow \dfrac{\sin^4 x+\cos^4 x}{\sin 2x}=\dfrac{1}{2}. (\dfrac{\sin^2 x +\cos^2 x}{\sin x \cos x})\Leftrightarrow \sin^4 x +\cos^4 x = 1 \Leftrightarrow 1-2sin^{2}x.cos^{2}x=1\Leftrightarrow sin^{2}2x=0
Điều kiện \sin 2x \ne 0 \Leftrightarrow x\ne \dfrac{k\pi}{2};\ k\in ZPT \Leftrightarrow \dfrac{\sin^4 x+\cos^4 x}{\sin 2x}=\dfrac{1}{2}. (\dfrac{\sin^2 x +\cos^2 x}{\sin x \cos x})\Leftrightarrow \sin^4 x +\cos^4 x = 1 \Leftrightarrow 1-2sin^{2}x.cos^{2}x=1\Leftrightarrow sin^{2}2x=0\Leftrightarrow sin2x=0