Điều kiện $\sin 2x \ne 0 \Leftrightarrow x\ne \dfrac{k\pi}{2};\ k\in Z$
PT$ \Leftrightarrow \dfrac{\sin^4 x+\cos^4 x}{\sin 2x}=\dfrac{1}{2}. (\dfrac{\sin^2 x +\cos^2 x}{\sin x \cos x})$
$\Leftrightarrow \sin^4 x +\cos^4 x = 1$
$\Leftrightarrow (1-\cos^2 x)^2 = (1-\cos^2 x)(1+\cos^2 x)$
$\Leftrightarrow (1-\cos^2 x) .[1-\cos^2 x-1-\cos^2 x]=0$ dễ rồi tự làm đi