Đáp án:
$S=\{0\}$
Giải thích các bước giải:
$\sin\left(x+\dfrac{\pi}{6}\right)=\dfrac{1}{2}$
$⇔\left[ \begin{array}{l}x+\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\x+\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi\end{array} \right.⇔\left[ \begin{array}{l}x=k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
$x\in\left(-\dfrac{\pi}{2};\dfrac{2\pi}{3}\right)$
TH1: $x=k2\pi$
$⇒-\dfrac{\pi}{2}<k2\pi<\dfrac{2\pi}{3}$
$⇒-\dfrac{1}{4}<k<\dfrac{1}{3}$
$k\in\mathbb Z⇒k=0$
$⇒x=0$
TH2: $x=\dfrac{2\pi}{3}+k2\pi$
$⇒-\dfrac{\pi}{2}<\dfrac{2\pi}{3}+k2\pi<\dfrac{2\pi}{3}$
$⇒-\dfrac{7}{12}<k<0$
$⇒k\in\varnothing$
Vậy $S=\{0\}$.
