Question

giải pt : \(2cos\left(\dfrac{2x}{3}+\dfrac{\pi}{6}\right)+\sqrt{3}=0\)

Answer 1

\(\Leftrightarrow cos\left(\dfrac{2x}{3}+\dfrac{\pi}{6}\right)=-\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2x}{3}+\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi\\\dfrac{2x}{3}+\dfrac{\pi}{6}=-\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k3\pi\\x=-\dfrac{3\pi}{2}+k3\pi\end{matrix}\right.\)