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