a: cos2x=cosx
=>\(\left[\begin{array}{l}2x=x+k2\pi\\ 2x=-x+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=k2\pi\\ 3x=k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=k2\pi\\ x=\frac{k2\pi}{3}\end{array}\right.\)
=>\(x=\frac{k2\pi}{3}\)
p: \(cos2x=\sqrt3\left(\sin2x-1\right)\)
=>\(\sqrt3\cdot\sin2x-\sqrt3-cos2x=0\)
=>\(\frac{\sqrt3}{2}\cdot\sin2x-\frac12\cdot cos2x=\frac{\sqrt3}{2}\)
=>\(\sin\left(2x-\frac{\pi}{6}\right)=\frac{\sqrt3}{2}\)
=>\(\left[\begin{array}{l}2x-\frac{\pi}{6}=\frac{\pi}{3}+k2\pi\\ 2x-\frac{\pi}{6}=\pi-\frac{\pi}{3}+k2\pi=\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac{\pi}{2}+k2\pi\\ 2x=\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{4}+k\pi\\ x=\frac{5}{12}\pi+k\pi\end{array}\right.\)
