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?

The solution is much more direct  if you think in terms of congruences:
\begin{align}
\cos 4x =\cos 2x&\iff 4x\equiv \pm2x \mod 2\pi\iff\begin{cases}2x\equiv 0\pmod{2\pi},\\ 6x\equiv 0\pmod{2\pi},\end{cases}\\
&\iff\begin{cases}x\equiv 0\pmod{\pi},\\[1ex] x\equiv 0\mod{\Bigl(\dfrac\pi3\Bigr)}.\end{cases}
\end{align}
