Đáp án:
Giải thích các bước giải:
`cos\ 2x+3sin\ x=2`
`⇔ 1-2sin^2 x+3sin\ x-2=0`
`⇔ 2sin^2 x-3sin\ x+1=0`
`⇔ (sin\ x-1)(2sin\ x-1)=0`
`⇔` \(\left[ \begin{array}{l}sin\ x=1\\sin\ x=\dfrac{1}{2}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\\sin\ x=sin\ (\dfrac{\pi}{6})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={\frac{\pi}{2}+k2\pi\ (k \in \mathbb{Z});\frac{\pi}{6}+k2\pi\ (k \in \mathbb{Z});\frac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})}`
