Giải phương trình \(\tan(3x-1)=-\dfrac{\sqrt{3}}{3}\).
\(x=\dfrac{1}{3}+\dfrac{5\pi}{18}+k\dfrac{\pi}{3}\,(k\in\mathbb{Z})\) | |
\(x=\dfrac{1}{3}+\dfrac{\pi}{18}+k\dfrac{\pi}{3}\,(k\in\mathbb{Z})\) | |
\(x=\dfrac{5\pi}{18}+k\dfrac{\pi}{3}\,(k\in\mathbb{Z})\) | |
\(x=\dfrac{1}{3}-\dfrac{\pi}{6}+k\pi\,(k\in\mathbb{Z})\) |
Chọn phương án A.
\(\begin{aligned}
\tan(3x-1)=-\dfrac{\sqrt{3}}{3}\Leftrightarrow&3x-1=\dfrac{5\pi}{6}+k\pi\\
\Leftrightarrow&3x=1+\dfrac{5\pi}{6}+k\pi\\
\Leftrightarrow&x=\dfrac{1}{3}+\dfrac{5\pi}{18}+k\dfrac{\pi}{3}\,(k\in\mathbb{Z})
\end{aligned}\)