Solving $\sin(5x) = \sin(x)$ If I have an equation:

$$\sin(5x) = \sin(x)$$

In what case can I equate $$5x = x$$

Is it only when there is a multiply of $2\pi n$ on either side, where n is any integer so

$$ 5x = x+2\pi n$$

Also with this method can I get every possible solution or does that not work?

I know I can use the sine addition formula but I want to see others I way I can solve this.
You might find this useful:

$$
\sin(A)=\sin(B) \iff A=B +n2\pi\ \mbox{ or }\ A=\pi-B +n2\pi,
$$
with $n\in\mathbb{Z}$. 
$$
\cos(A)=\cos(B) \iff A=B +n2\pi\ \mbox{ or }\ A=-B +n2\pi,
$$
with $n\in\mathbb{Z}$. 

So you need to solve two equations separately:
$$5x=x+n2\pi$$
and 
$$5x=\pi-x+n2\pi.$$
Good luck :)



Edit:
One way to prove 
$$
\sin(x)=\sin(y) \iff x=y +n2\pi\ \mbox{ or }\ x=\pi-y +n2\pi,
$$
is to just verify the solutions work, using a consequence of Euler's formula:
$$\sin(x)=\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}.$$
So, here we go:
$$
\sin(x)=\sin(y) \iff \frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}=\frac{\mathrm{e}^{iy}-\mathrm{e}^{-iy}}{2i} \iff \mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{iy}-\mathrm{e}^{-iy}.
$$


Now note that $x=y+n2\pi$ is indeed a solution:
$$\mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{i(y+n2\pi)}-\mathrm{e}^{-i(y+n2\pi)}=\mathrm{e}^{iy}\mathrm{e}^{in2\pi}-\mathrm{e}^{-iy}\mathrm{e}^{-in2\pi}=\mathrm{e}^{iy}-\mathrm{e}^{-iy}
$$
and that $x=\pi-y+n2\pi$ is indeed a solution:
$$\mathrm{e}^{ix}-\mathrm{e}^{-ix}=\mathrm{e}^{i(\pi-y+n2\pi)}-\mathrm{e}^{-i(\pi-y+n2\pi)}=\mathrm{e}^{-iy}\mathrm{e}^{i(2n+1)\pi}-\mathrm{e}^{iy}\mathrm{e}^{-i(2n+1)\pi}=\mathrm{e}^{iy}-\mathrm{e}^{-iy}.$$


The last equalities use $\mathrm{e}^{in\pi}$ is equal to $1$ if $n$ is even and equal to $-1$ if $n$ is odd.
Alternative solution: expand the LHS:
$$16\sin^5 x - 20\sin^3 x + 5\sin x = \sin(5x) = \sin x,$$
and $s = \sin x$ is a solution of
$$(16s^4 - 20s^2 + 4)s = 16s^5 - 20s^3 + 4s = 0.$$