Answer
$\left\{0, \frac{2\pi}{3}, \frac{4\pi}{3}\right\}$
Work Step by Step
Subtract $\cos{x}$ to both sides of the equation:
$0=\cos{(2x)}-\cos{x}$
RECALL:
$\cos{(2x)}=2\cos^2{x}-1$
Use the identity above to obtain:
$0=\cos{(2x)} -\cos{x}
\\0=(2\cos^2{x}-1)-\cos{x}
\\0=2\cos^2{x}-\cos{x}-1$
Factor the trinomial to obtain:
$0=(2\cos{x}+1)(\cos{x}-1)$
Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
&2\cos{x}+1=0 &\text{or} &\cos{x}-1=0
\\&2\cos{x}=-1 &\text{or} &\cos{x}=1
\\&\cos{x}=-\frac{1}{2} &\text{or} &\cos{x}=1
\\&x=\cos^{-1}{(-\frac{1}{2})} &\text{or} &x=\cos^{-1}{(1)}
\end{array}
Use a calculator's inverse cosine function to obtain:
\begin{array}{ccc}
&x=\frac{2\pi}{3} &\text{or} &x=0
\end{array}
Note that for an angle $x$ in Quadrant I, $\cos{x} = \cos{(2\pi-x)}$.
Thus, if $\frac{2\pi}{3}$ is a solution, then $(2\pi-\frac{2\pi}{3})=\frac{4\pi}{3}$ is also a solution.
Therefore, the solution set is:
$\left\{0, \frac{2\pi}{3}, \frac{4\pi}{3}\right\}$
