## Precalculus (6th Edition) Blitzer

The solutions of the provided trigonometric equation are $\theta =\frac{\pi }{3},\frac{5\pi }{3},0$.
We have the equation as $2+\cos 2\theta =3\cos \theta$. By solving the equation, \begin{align} & 2+\cos 2\theta =3\cos \theta \\ & 2+2co{{s}^{2}}\theta -1=3\cos \theta \\ & 2{{\cos }^{2}}\theta -3\cos \theta +1=0 \end{align} Then, \begin{align} & 2{{\cos }^{2}}\theta -2\cos \theta -\cos \theta +1=0 \\ & \left( 2\cos \theta -1 \right)(\cos \theta -1)=0 \\ & \cos \theta =\frac{1}{2},1 \end{align} Hence, in the provided range, we get, $\theta =\frac{\pi }{3},\frac{5\pi }{3},0$