\(cos^2x-\sqrt{3}sin2\text{x}=1+sin^2x\\ \Leftrightarrow cos2x-\sqrt{3}sin2\text{x}=1\\ \Leftrightarrow\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2\text{x}=\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos2x-sin\frac{\pi}{3}\cdot sin2\text{x}=\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos2x-sin\frac{\pi}{3}\cdot sin2\text{x}=\frac{1}{2}\\ \Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=cos\frac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=\frac{\pi}{3}+a2\pi\\2x+\frac{\pi}{3}=-\frac{\pi}{3}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=a\pi\\x=-\frac{\pi}{3}+b\pi\end{matrix}\right.\)
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