ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}}=1+sin2x\)
\(\Leftrightarrow\frac{cosx-sinx}{cosx+sinx}=1+sin2x\)
\(\Leftrightarrow cosx-sinx=\left(cosx+sinx\right)\left(1+sin2x\right)\)
\(\Leftrightarrow cosx.sin2x+2sinx+sinx.sin2x=0\)
\(\Leftrightarrow cos^2x.sinx+sinx+sin^2x.cosx=0\)
\(\Leftrightarrow sinx\left(cos^2x+sinx.cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\cos^2x+sinx.cosx+1=0\left(1\right)\end{matrix}\right.\)
Xét (1), chia 2 vế cho \(cos^2x\) ta được:
\(1+tanx+1+tan^2x=0\)
\(\Leftrightarrow tan^2x+tanx+2=0\) (vô nghiệm)
