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