Đáp án:
a) $S=\left\{\dfrac{5\pi}{3}+k4\pi;-\dfrac{5\pi}{3}+k4\pi\,\bigg{|}\,k\in\mathbb Z\right\}$
b) $S=\{\pi+k4\pi\,\bigg{|}\,k\in\mathbb Z\}$
Giải thích các bước giải:
a) $\cos\dfrac{x}{2}=-\dfrac{\sqrt 3}{2}$
$⇔\left[ \begin{array}{l}\dfrac{x}{2}=\dfrac{5\pi}{6}+k2\pi\\\dfrac{x}{2}=-\dfrac{5\pi}{6}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
$⇔\left[ \begin{array}{l}x=\dfrac{5\pi}{3}+k4\pi\\x=-\dfrac{5\pi}{3}+k4\pi\end{array} \right.\,\,(k\in\mathbb Z)$
Vậy $S=\left\{\dfrac{5\pi}{3}+k4\pi;-\dfrac{5\pi}{3}+k4\pi\,\bigg{|}\,k\in\mathbb Z\right\}$.
b) $\sin\dfrac{x}{2}=1$
$⇔\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\,\,(k\in\mathbb Z)$
$⇔x=\pi+k4\pi\,\,(k\in\mathbb Z)$
Vậy $S=\{\pi+k4\pi\,\bigg{|}\,k\in\mathbb Z\}$.
