General solution of the equation  $1+\cos 3\theta =2\cos 2\theta$ is

$\begin{array}{l}1+\cos 3\theta =2\cos 2\theta \\ \Rightarrow 1+4{\cos }^3\theta -3\cos \theta =2(2{\cos }^2\theta -1)\\ \Rightarrow 4{\cos }^3\theta -4{\cos }^2\theta -3\cos \theta +3=0\\ \Rightarrow (\cos \theta -1)(4{\cos }^2\theta -3)=0\\ \Rightarrow \cos \theta =1or{\cos }^2\theta =\frac{3}{4}={\left(\frac{\sqrt{3}}{2}\right)}^2={\cos }^2\left(\frac{\pi }{6}\right)\\ \Rightarrow \theta =2n\pi or\theta =n\pi \pm \frac{\pi }{6},n\in z\end{array}$