Đk: \(cosx\ne0\Leftrightarrow x\ne\frac{\pi}{2}+k\pi\) (k thuộc Z)
<=> tanx(2sinx-1)=0
<=> \(\left[{}\begin{matrix}tanx=0\\sinx=\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)