General solution of $\cos{4x}=\cos{2x}$ My attempt:
\begin{align}
\cos{4x}-\cos{2x} &=0 \\
\cos^2{2x}-\sin^2{2x}-\cos{2x} &=0 \\
\cos^2{2x}-1+\cos^2{2x}-\cos{2x} &=0 \\
\cos^2{2x}-\cos{2x}-1 &= 0 \\
(2\cos{2x}+1)(\cos{2x}-1) &= 0,
\end{align}
which leads to
$$\cos2x= -\frac{1}{2} \hspace{5mm} \text{or} \hspace{5mm} \cos2x=1,$$
$$2x=2n\pi\pm{2\pi\over3} \hspace{5mm} \text{or} \hspace{5mm} 2x=2n\pi$$
and
$$x=n\pi\pm{\pi\over3} \hspace{5mm} \text{  or  } \hspace{5mm} x=n\pi.$$

Book Solution:
\begin{align}
\cos{4x}-\cos{2x} &=0 \\
-2\sin{3x}\sin{x} &=0 \\
\sin{3x}\sin{x} &=0.
\end{align}
$$\sin{3x}=0 \hspace{5mm} \text{or} \hspace{5mm} \sin{x}=0$$
$$3x=n\pi \hspace{5mm} \text{or} \hspace{5mm} x=n\pi$$
$$x=\frac{n\pi}{3} \hspace{5mm} \text{or} \hspace{5mm} x=n\pi$$
My question is: How can there be two general solutions of a same equation Or did I made a mistake somewhere?

They are  same answers.
I think it's better the following way.
$$4x=2x+2\pi k$$ or
$$4x=-2x+2\pi k,$$
where $k\in\mathbb Z$, which gives else the same answer:
$$\left\{\frac{\pi k}{3}|k\in\mathbb Z\right\}$$
