SOLUTION
From
[tex]\begin{gathered} \cos 2x=\cos x \\ \cos ^2x-\sin ^2x=\cos x \\ \cos ^2x-(1-\cos ^2x)=\cos x \\ 2\cos ^2x-1=\cos x \\ 2\cos ^2x-\cos x-1=0 \\ \text{From the quadratic formula} \\ \cos x=\frac{1\pm\sqrt[]{1-(-8)}}{4} \\ \\ \cos x=\frac{1\pm3}{4} \\ \cos x=\text{ 1 or -}\frac{1}{2} \\ \text{Taking the cos}^{-1}of\text{ 1 and -}\frac{1}{2} \\ We\text{ have }\theta\text{ = 0, }\frac{2\pi}{3},\frac{4\pi}{3},\frac{8\pi}{3}\ldots\ldots\ldots2\pi \end{gathered}[/tex]
So your answer is
[tex]0,\text{ }\frac{2\pi}{3},\text{ }\frac{4\pi}{3}[/tex]
