Đáp án:
$S=\left\{\dfrac{\pi}{4}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
$\dfrac{1}{2}\sin x+\dfrac{1}{2}\cos x=\dfrac{\sqrt 2}{2}$
$⇔\dfrac{\sqrt 2}{2}\sin x+\dfrac{\sqrt 2}{2}\cos x=1$
$⇔\sin x.\cos\dfrac{\pi}{4}+\cos x.\sin\dfrac{\pi}{4}=1$
$⇔\sin\left(x+\dfrac{\pi}{4}\right)=1$
$⇔x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\,\,(k\in\mathbb Z)$
$⇔x=\dfrac{\pi}{4}+k2\pi$
Vậy $S=\left\{\dfrac{\pi}{4}+k2\pi\,\bigg{|}\,k\in\mathbb Z\right\}$.
