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