ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(cosx+\dfrac{sinx}{cosx}=1+\dfrac{sin^2x}{cosx}\)
\(\Rightarrow cos^2x+sinx=cosx+sin^2x\)
\(\Leftrightarrow cos^2x-sin^2x+sinx-cosx=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx\right)-\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\\cosx+sinx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\left(loại\right)\end{matrix}\right.\)
