ĐKXĐ: ...
\(\frac{cosx}{sinx}-\frac{sinx}{cosx}=2cos2x.sinx\Leftrightarrow\frac{cos^2x-sin^2x}{sinx.cosx}=2cos2x.sinx\)
\(\Leftrightarrow\frac{cos2x}{sinx.cosx}=2cos2x.sinx\Leftrightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\\frac{1}{sinx.cosx}=2sinx\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2sin^2x.cosx=1\Leftrightarrow2cosx\left(1-cos^2x\right)=1\)
\(\Leftrightarrow2cos^3x-2cosx+1=0\)
\(\Leftrightarrow\left(2cos^3x-2cos^2x+\frac{4\sqrt{3}}{9}\right)+\frac{9-4\sqrt{3}}{9}=0\)
\(\Leftrightarrow2\left(cosx-\frac{1}{\sqrt{3}}\right)^2\left(cosx+\frac{2}{\sqrt{3}}\right)+\frac{9-4\sqrt{3}}{9}=0\)
Vế trái luôn dương nên pt (1) vô nghiệm
