ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\frac{1}{2}\left[1-cos\left(x-\frac{\pi}{2}\right)\right].\frac{sin^2x}{cos^2x}-\frac{1}{2}\left(1+cosx\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\frac{1-cos^2x}{1-sin^2x}-\left(1+cosx\right)=0\)
\(\Leftrightarrow\frac{\left(1-sinx\right)\left(1-cosx\right)\left(1+cosx\right)}{\left(1-sinx\right)\left(1+sinx\right)}-\left(1+cosx\right)=0\)
\(\Leftrightarrow\left(1+cosx\right)\left(\frac{1-cosx}{1+sinx}-1\right)=0\)
\(\Leftrightarrow\frac{\left(1+cosx\right)\left(-cosx-sinx\right)}{1+sinx}=0\)
\(\Leftrightarrow\left(1+cosx\right)sin\left(x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
