Phương trình \(\sqrt{3}\sin x-\cos x=2\) có nghiệm là
| \(x=\dfrac{2\pi}{3}+k\dfrac{2\pi}{3},\,k\in\mathbb{Z}\) | |
| \(x=\dfrac{2\pi}{3}+k\pi,\,k\in\mathbb{Z}\) | |
| \(x=\dfrac{\pi}{3}+k2\pi,\,k\in\mathbb{Z}\) | |
| \(x=\dfrac{2\pi}{3}+k2\pi,\,k\in\mathbb{Z}\) |
Chọn phương án D.
\(\begin{aligned}
\sqrt{3}\sin x-\cos x=2\Leftrightarrow&\sin\left(x-\dfrac{\pi}{6}\right)=1\\
\Leftrightarrow&x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\\
\Leftrightarrow&x=\dfrac{2\pi}{3}+k2\pi,\,k\in\mathbb{Z}
\end{aligned}\)