Đáp án:
\(\left[ \begin{array}{l}
x = - \dfrac{\pi }{4}\\
x = \dfrac{{5\pi }}{{12}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
2\sin \left( {2x + \dfrac{\pi }{3}} \right) + 1 = 0\\
\to \sin \left( {2x + \dfrac{\pi }{3}} \right) = - \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
2x + \dfrac{\pi }{3} = - \dfrac{\pi }{6} + k2\pi \\
2x + \dfrac{\pi }{3} = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{\pi }{4} + k\pi \\
x = \dfrac{{5\pi }}{{12}} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
Do:x < \dfrac{\pi }{2}\\
\to \left[ \begin{array}{l}
x = - \dfrac{\pi }{4}\\
x = \dfrac{{5\pi }}{{12}}
\end{array} \right.
\end{array}\)
