Đáp án:
Giải thích các bước giải:
`3cos^2\ 2x-3sin^2x+cos^2x=0`
`⇔ 3cos^2\ 2x-3(\frac{1-cos\ 2x}{2})+\frac{1+cos\ 2x}{2}=0`
`⇔ 6cos^2\ 2x-3(1-cos\ 2x)+1+cos\ 2x=0`
`⇔ 6cos^2\ 2x-3+3cos\ 2x+1+cos\ 2x=0`
`⇔ 6cos^2\ 2x+4cos\ 2x-2=0`
`⇔` \(\left[ \begin{array}{l}cos\ 2x=\dfrac{1}{3}\\cos\ 2x=-1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x=\pm arccos\ \Bigg(\dfrac{1}{3}\Bigg)+k2\pi\ (k \in \mathbb{Z})\\2x=\pi+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\pm \dfrac{arccos\ \Bigg(\dfrac{1}{3}\Bigg)}{2}+k\pi\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{2}+k\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={\pm \frac{arccos\ (\frac{1}{3})}{2}+k\pi\ (k \in \mathbb{Z});x=\frac{\pi}{2}+k\pi\ (k \in \mathbb{Z})}`
