Đáp án:
$S=\left\{-\dfrac{\pi}{6}+k2\pi;\dfrac{11\pi}{18}+k\dfrac{2\pi}{3}\,\bigg|\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
$\sin\left(2x-\dfrac{\pi}{3}\right)=-\cos x$
$⇔\sin\left(2x-\dfrac{\pi}{3}\right)=-\sin\left(\dfrac{\pi}{2}-x\right)$
$⇔\sin\left(2x-\dfrac{\pi}{3}\right)=\sin\left(x-\dfrac{\pi}{2}\right)$
$⇔\left[ \begin{array}{l}2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{2}+k2\pi\\2x-\dfrac{\pi}{3}=\dfrac{3\pi}{2}-x+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
$⇔\left[ \begin{array}{l}x=-\dfrac{\pi}{6}+k2\pi\\3x=\dfrac{11\pi}{6}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
$⇔\left[ \begin{array}{l}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{11\pi}{18}+k\dfrac{2\pi}{3}\end{array} \right.\,\,(k\in\mathbb Z)$
Vậy $S=\left\{-\dfrac{\pi}{6}+k2\pi;\dfrac{11\pi}{18}+k\dfrac{2\pi}{3}\,\bigg|\,k\in\mathbb Z\right\}$.
