ĐKXĐ: ....
\(\Leftrightarrow\frac{1-2\sqrt{2}\left(sin2x+cos2x\right)}{sin4x}=\frac{6sin^2\left(x-\frac{\pi}{8}\right)}{cos^2\left(x-\frac{\pi}{8}\right)}\)
\(\Leftrightarrow\frac{1-2\sqrt{2}\left(sin2x+cos2x\right)}{sin4x}=\frac{6\left(1-cos\left(2x-\frac{\pi}{4}\right)\right)}{1+cos\left(2x-\frac{\pi}{4}\right)}\)
\(\Leftrightarrow\frac{1-2\sqrt{2}\left(sin2x+cos2x\right)}{sin4x}=\frac{6\left(\sqrt{2}-\left(sin2x+cos2x\right)\right)}{\sqrt{2}+sin2x+cos2x}\)
Đặt \(sin2x+cos2x=a\Rightarrow sin4x=a^2-1\)
\(\frac{1-2\sqrt{2}a}{a^2-1}=\frac{6\sqrt{2}-6a}{\sqrt{2}+a}\Leftrightarrow6a^3-8\sqrt{2}a^2-9a+7\sqrt{2}=0\)
\(\Leftrightarrow\left(2a-\sqrt{2}\right)\left(6a^2-5\sqrt{2}a-14\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=\frac{\sqrt{2}}{2}\\6a^2-5\sqrt{2}a-14=0\end{matrix}\right.\)
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