Đáp án:
\(\left[ \begin{array}{l}
x = - \frac{\pi }{{12}} - \frac{1}{2} + k\pi \\
x = \frac{{7\pi }}{{12}} - \frac{1}{2} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
$\begin{array}{l}
\sin \left( {2x + 1} \right) = - \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l}
2x + 1 = - \frac{\pi }{6} + k2\pi \\
2x + 1 = \frac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = - \frac{\pi }{6} - 1 + k2\pi \\
2x = \frac{{7\pi }}{6} - 1 + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - \frac{\pi }{{12}} - \frac{1}{2} + k\pi \\
x = \frac{{7\pi }}{{12}} - \frac{1}{2} + k\pi
\end{array} \right.
\end{array}$
