How can one quickly find out the number of solutions to $$\cos^2{x}=\cos{2x}, \ \ \ 0 \leq x \leq 2\pi \ ? $$
I rewrote the equation as $$\cos2x=\cos^2{x}-\sin^2{x} \Longleftrightarrow -\sin^2{x} = 0 \Longleftrightarrow \sin{x}=0.$$
So, the equation $\sin{x}=0$ has roots on $0, \pi$ and $2\pi$. So the answer should be 3. Is this an acceptable and all correct reasoning or can I improve on any detail? Is the answer correct?
