ĐKXĐ: \(cosx\ne0\Rightarrow x\ne\dfrac{\pi}{2}+k\pi\)
\(\dfrac{tan^2x+tanx}{tan^2x+1}=\dfrac{\sqrt{2}}{2}sin\left(\dfrac{\pi}{4}+x\right)\)
\(\Leftrightarrow cos^2x\left(tan^2x+tanx\right)=\dfrac{\sqrt{2}}{2}\left(sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx\right)\)
\(\Leftrightarrow sin^2x+sinxcosx=\dfrac{1}{2}\left(sinx+cosx\right)\)
\(\Leftrightarrow sinx\left(sinx+cosx\right)-\dfrac{1}{2}\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-\dfrac{1}{2}\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx+cosx=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\\sqrt{2}.sin\left(x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=\dfrac{-\pi}{4}+k\pi\end{matrix}\right.\)
