How to solve trig equations and get all the solutions using graphs, $\cos(2x-\pi/3)=\cos(x)$ The question is to solve 
$$\cos\left(2x-\frac{\pi}{3}\right)=\cos(x)$$
I originally approached this using the addition formulae but the mark scheme showed a way by first replacing $x$ on the right with $2\pi-x$ and I understand this is due to the $\cos $ graph, however I don't understand where all the solutions came from, would really appreciate help understanding 
Answers: $\frac{\pi}{3}, \frac{7\pi}{3}$ and $\frac{13\pi}{3}$.
Thanks 
 A: Note that because of the $2\pi$-periodicity and evenness of the cosine function, we have $\cos(x)=\cos(\pm x+2k\pi)$ for all integer $k$.  
If $\cos(2x-\pi/3)=\cos(x)$, then $2x-\pi/3=\pm x+2k\pi$, whereupon we find
$$x=\frac{\pi/3+2k\pi}{2\pm 1}$$
A: $$\cos { \left( 2x-\frac { \pi  }{ 3 }  \right) =\cos { x }  } $$
$$2x-\frac { \pi  }{ 3 } =x+2\pi n,\quad \quad \Rightarrow x=\frac { \pi  }{ 3 } +2\pi n,\quad $$
$$2x-\frac { \pi  }{ 3 } =-x+2\pi n,n=0,\pm 1,\pm 2,..\quad \Rightarrow x=\frac { \pi  }{ 9 } +\frac { 2\pi n }{ 3 } ,n=0,\pm 1\quad ,\pm 2,..$$
A: There are other solutions. To see this, it's simpler to use congruences: the basic equations in trigonometry are


*

*$\cos a=\cos b\iff a\equiv \pm b\mod 2\pi$,

*$\sin a=\sin b\iff a\equiv\begin{cases}b\\\pi-b\end{cases}\mod 2\pi,$

*$\tan a=\tan b\iff a\equiv b\mod \pi$.


Here you obtain
$$2x-\frac\pi3\equiv\pm x\mod2\pi\iff\begin{cases}x\equiv \dfrac\pi3\mod 2\pi\\[1ex]3x\equiv \dfrac\pi3\mod2\pi\end{cases}\iff\begin{cases}x\equiv \dfrac\pi3\mod 2\pi\\[1ex]x\equiv \dfrac\pi9\mod\dfrac{2\pi}3\end{cases}$$
