Đáp án:
(sinx + 1)(2cosx -1 )=0
⇔ \(\left[ \begin{array}{l}sinx+1 =0\\2cosx -1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}sinx = -1\\cosx=\frac{1}{2}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{-\pi}{2} + k2\pi\\x=±\frac{\pi}{3} + k2\pi\end{array} \right.\)