Алгебра: Прошу помочь с тригонометрией. По возможности с подробным решением

$\left(\sqrt{3}\sin2x+\cos2x\right)^2=8-4\cos\left(2x+\dfrac{2\pi}{3}\right);\medskip\\\left(\dfrac{\sqrt{3}\sin2x+\cos2x}{2}\right)^2=2-\cos\left(2x+\dfrac{2\pi}{3}\right);\medskip\\\left(\dfrac{\sqrt{3}}{2}\sin2x+\dfrac{1}{2}\cos2x\right)^2=2-\cos\left(2x+\dfrac{2\pi}{3}\right);\medskip\\\left(\sin\dfrac{\pi}{3}\sin2x+\cos\dfrac{\pi}{3}\cos2x\right)^2=2-\cos\left(2x+\pi-\dfrac{\pi}{3}\right);\medskip\\\cos^2\left(2x-\dfrac{\pi}{3}\right)=2-\cos\left(\pi+\left(2x-\dfrac{\pi}{3}\right)\right);$

$\cos^2\left(2x-\dfrac{\pi}{3}\right)=2+\cos\left(2x-\dfrac{\pi}{3}\right);$

Пусть $\cos\left(2x-\dfrac{\pi}{3}\right)=y$ . Тогда,

$y^2=2+y\,;\medskip\\y^2-y-2=0\,;\medskip\\\begin{cases}\left[\begin{gathered}y=-1\\y=2\end{gathered}\\y\in\left[-1;1\right]\end{cases}\!\!\!\!\Leftrightarrow y=-1$

$\cos\left(2x-\dfrac{\pi}{3}\right)=-1\,;\medskip\\2x-\dfrac{\pi}{3}=\pi+2\pi n,\, n\in\mathbb{Z}\,;\medskip\\x=\dfrac{\pi}{2}+\dfrac{\pi}{6}+\pi n,\, n\in\mathbb{Z}\,;\medskip\\x=\pi n+\dfrac{2\pi}{3},\, n\in\mathbb{Z}\,.$

Ответ. $x=\pi n+\dfrac{2\pi}{3},\, n\in\mathbb{Z}\,.$