Solving the equation: $\cos(x)= \cos(2x)$ I'll be glad if someone could explain the justification of this solution:
$$\cos(x)=\cos(2x),\; [0^{\circ},360^{\circ})$$
$$\Rightarrow x=\pm2x+360^{\circ}k,\; k\in\mathbb{Z}\Rightarrow x=0^{\circ}, 120^{\circ}, 240^{\circ}$$
How come I can cancel the "cos" like that?
My solution is:
$$\cos(2x)-\cos(x)=0 \Rightarrow (2\cos^{2}x-1)-\cos(x)=0 \Rightarrow \cos x=1 ,\; \cos x=-\frac{1}{2} $$
Then: $x=0^{\circ}, 120^{\circ}, 240^{\circ}$
Thanks.
 A: $\cos x = \cos y$ when and only when $x = y + 360 k$ or $x = -y + 360 k$.  Use these two relations to obtain the first method.  This follows from the definition of $\cos x$ being the horizontal component of an angle $x$ on the unit circle.  Only (at most) two points have the same horizontal component, namely angle $x$ and $-x$.
A: Hint:
Use $\cos (2x)= 2 \cos^2(x)-1$.
$2 \cos^2(x)-1=\cos(x)$
Take $\cos(x)=y$
$2y^2-1-y=0 \implies (2y+1)(y-1)=0$
Justification for your effort:
$\cos(x)=\cos(2x)$
$\cos(x)= \cos(360^\circ \pm 2x)$,
You know that $x \neq 2x$ (Except for $x=0$). So, you must have $x=360 \pm 2x$. 
A: We know $$\cos(A-B)-\cos(A+B)=2\sin A\sin B$$
For $\cos x=\cos 2x,x=A-B,A+B=2x\implies A=\frac{3x}2, B=\frac x2$
So, $\cos x-\cos 2x=0\implies 2\sin \frac{3x}2\sin \frac x2=0 $
If $\sin \frac x2=0, \frac x2=n180^\circ\implies x=n 360^\circ$ where $n$ is any integer
If $\sin \frac{3x}2=0, \frac{3x}2=m180^\circ\implies x=m120^\circ$ where $m$ is any integer
Observe that the second set is a super-set of the first
So, we need $0\le 120m<360\iff 0\le m<3$
