ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\dfrac{2\left(sin^2x-cos^2x\right)}{sinx.cosx}=\dfrac{sin4x}{\sqrt{3}cos2x+2}\)
\(\Rightarrow\dfrac{-4cos2x}{sin2x}=\dfrac{2sin2x.cos2x}{\sqrt{3}cos2x+2}\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\-\dfrac{2}{sin2x}=\dfrac{sin2x}{\sqrt{3}cos2x+2}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow2\sqrt{3}cos2x+4=-sin^22x\)
\(\Leftrightarrow2\sqrt{3}cos2x+4=cos^22x-1\)
\(\Leftrightarrow cos^22x-2\sqrt{3}cos2x-5=0\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=\sqrt{3}+2\sqrt{2}>1\left(loại\right)\\cos2x=\sqrt{3}-2\sqrt{2}< -1\left(loại\right)\end{matrix}\right.\)
