Đáp án:
\(S = \{0\}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \sin\left(x +\dfrac{\pi}{6}\right) = \dfrac12\\
\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{6} = \dfrac{\pi}{6} + k2\pi\\x + \dfrac{\pi}{6} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = k2\pi\\x = \dfrac{2\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Ta có:}\\
\quad - \dfrac{\pi}{2} < x < \dfrac{2\pi}{3}\\
\Leftrightarrow \left[\begin{array}{l}-\dfrac{\pi}{2} < k2\pi < \dfrac{2\pi}{3}\\-\dfrac{\pi}{2} < \dfrac{2\pi}{3} +k2\pi < \dfrac{2\pi}{3}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}-\dfrac14 < k < \dfrac13\\-\dfrac{7}{12} < k < 0\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}k = 0\\k \in\varnothing\end{array}\right.\\
\Rightarrow x = 0\\
\text{Vậy}\ S = \{0\}
\end{array}\)
