\(\sqrt{3}cos2x+sin2x=-\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}\left(2cos^2x-1+1\right)+sin2x=0\)
\(\Leftrightarrow2\sqrt{3}cos^2x+2sinx.cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}cosx+sinx\right)=0\)
\(\Leftrightarrow cosx\left(\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\right)=0\)
\(\Leftrightarrow cosx.cos\left(x-\dfrac{\pi}{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos\left(x-\dfrac{\pi}{6}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{2\pi}{3}+k\pi\end{matrix}\right.\)
⇔ \(\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x=\dfrac{-\sqrt{3}}{2}\)
⇔ \(cos\left(2x-\dfrac{\pi}{6}\right)=cos\dfrac{5\pi}{6}\)
⇔ \(\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi\\2x-\dfrac{\pi}{6}=-\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\) với k là số nguyên
