Đáp án:
Giải thích các bước giải:
`\sqrt{3}cos\ 3x-sin\ 3x=-1`
`⇔ \frac{\sqrt{3}}{\sqrt{(\sqrt{3})^2+1^2}}cos\ 3x-\frac{1}{\sqrt{(\sqrt{3})^2+1^2}}sin\ 3x=\frac{-1}{\sqrt{(\sqrt{3})^2+1^2}}`
`⇔ \frac{\sqrt{3}}{2}cos\ 3x-\frac{1}{2}sin\ 3x=\frac{-1}{2}`
`⇔ sin\ \frac{\pi}{3} . cos\ 3x-cos\ \frac{\pi}{3} . sin\ 3x=\frac{-1}{2}`
`⇔ sin\ (3x-\frac{\pi}{3})=-1/2`
`⇔ sin\ (3x-\frac{\pi}{3})=sin\ -\frac{\pi}{6}`
`⇔` \(\left[ \begin{array}{l}3x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\3x-\dfrac{\pi}{3}=\pi+\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{\pi}{18}+k\dfrac{2\pi}{3}\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
