Đáp án:
Giải thích các bước giải:
`\sqrt{3}sin\ 2x+2cos^2 x=3`
`⇔ \sqrt{3}sin\ 2x+2.(\frac{1+cos\ 2x}{2})=3`
`⇔ \sqrt{3}sin\ 2x+1+cos\ 2x=3`
`⇔ \sqrt{3}sin\ 2x+cos\ 2x=2`
`⇔ \frac{\sqrt{3}}{\sqrt{(\sqrt{3})^2+1^2}}sin\ 2x+\frac{1}{\sqrt{(\sqrt{3})^2+1^2}}cos\ 2x=\frac{2}{\sqrt{(\sqrt{3})^2+1^2}}`
`⇔ \frac{\sqrt{3}}{2}sin\ 2x+\frac{1}{2}cos\ 2x=1`
`⇔ cos\ \frac{\pi}{6}}sin\ 2x+sin\ \frac{\pi}{6}}cos\ 2x=1`
`⇔ sin\ (2x+\frac{\pi}{6})=1`
`⇔ 2x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})`
`⇔ 2x=\frac{\pi}{3}+k2\pi\ (k \in \mathbb{Z})`
`⇔ x=\frac{\pi}{6}+k\pi\ (k \in \mathbb{Z})`
Vậy `S={\frac{\pi}{6}+k\pi\ (k \in \mathbb{Z})}`
