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