Đáp án:
$S=\left\{-\dfrac{\pi}{4}+k2\pi;\dfrac{\pi}{12}+k\dfrac{2\pi}{3}\,\bigg{|}\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
$\cos\left(x-\dfrac{\pi}{4}\right)=\sin\left(2x+\dfrac{\pi}{2}\right)$
$⇔\cos\left(x-\dfrac{\pi}{4}\right)=\cos\left(-2x\right)$
$⇔\cos2x=\cos\left(x-\dfrac{\pi}{4}\right)$
$⇔\left[ \begin{array}{l}2x=x-\dfrac{\pi}{4}+k2\pi\\2x=\dfrac{\pi}{4}-x+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
$⇔\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k2\pi\\3x=\dfrac{\pi}{4}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
$⇔\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{12}+k\dfrac{2\pi}{3}\end{array} \right.\,\,(k\in\mathbb Z)$
Vậy $S=\left\{-\dfrac{\pi}{4}+k2\pi;\dfrac{\pi}{12}+k\dfrac{2\pi}{3}\,\bigg{|}\,k\in\mathbb Z\right\}$.
