Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}\left( {2sin2x - 2} \right)\left( {cox + 1} \right) = 0\\ \Leftrightarrow \left\{ \begin{array}{l}2\sin \,2x - 2 = 0\\\cos x\, + 1 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\sin \,2x = 1\\cos\,x = - 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}2x = \frac{\pi }{2} + k2\pi \\x = \pi + k2\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{\pi }{4} + k\pi \\x = \pi + k2\pi \end{array} \right.\end{array}\]
