\(\left(1+2sinx\right)cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left(1+2sinx\right)\left(cos2x-\sqrt{3}sin2x\right)=1\)
\(\Leftrightarrow cos2x-\sqrt{3}sin2x+2sinx.cos2x-2\sqrt{3}sinx.sin2x=1\)
\(\Leftrightarrow1-2sin^2x-2\sqrt{3}.sinx.cosx+2sinx.cos2x-2\sqrt{3}sinx.sin2x=1\)
\(\Leftrightarrow-2sinx\left(sinx+\sqrt{3}cosx-cos2x+\sqrt{3}sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sinx+\sqrt{3}cosx=cos2x-\sqrt{3}sin2x\left(1\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=cos\left(2x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
