\(\left(3tanx+\sqrt{3}\right)\left(2sinx-1\right)=0\Leftrightarrow\left[\begin{array}{nghiempt}tanx=\frac{-\sqrt{3}}{3}\\sinx=\frac{1}{2}\end{array}\right.\)
\(TH1:tanx=\frac{-\sqrt{3}}{3}\Leftrightarrow x=\frac{-\pi}{6}+k\pi\\ TH2:sinx=\frac{1}{2}\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{array}\right.\)
