Does $x\neq\frac{\pi}{z}$ imply $xn_1\neq\frac{\pi}{z}n_2$? Sorry for all the confusion. Heres what I'm asking:
If $x\in(-\pi,0)\cup(0,\pi)$ cannot be written as $\frac{\pi}{z}$ for any whole number $z$ can we say that any integer multiple of $x$ can never be written as an integer multiple of $\frac{\pi}{z}$ for any whole number $z$?

I want to know because I want to know if the sequence $(\cos(xn))_{n\in\mathbb{N}}$ with $x\in(-\pi,0)\cup(0,\pi)$ has any two elements that are equal if $x\neq\frac{\pi}{z}$ with $z\in\mathbb{Z}$
 A: It looks like there is some ambiguity here due to poor notation. 
What I think you mean is that say there is an $x\in\mathbb{R}$ such that there is no $z\in \mathbb{Z}$ such that $x=\frac{\pi}{z}$. In this case you have to worry about $x=\frac{p\pi}{q}$ with $p,q\in \mathbb{Z}$ (ie x is a rational number times $\pi$). 
(counter example is $x=\frac{2\pi}{3}$ there is no $z\in\mathbb{Z}$)
In your use case this works out to $\cos(3\frac{2\pi}{3})=\cos(6\frac{2\pi}{3})=1$

If you remove that option then I think by definition your statement is true. (ie $\nexists p,q\in\mathbb{Z}:x=\frac{p\pi}{q}\implies\nexists p,q,z\in\mathbb{Z}:x=\frac{p\pi}{q z}$
$\implies \nexists p,q\in\mathbb{N}, z\in \mathbb{Z}:q x=\frac{p\pi}{z}$
$\implies\nexists n_1,n_2\in \mathbb{N},z\in\mathbb{Z}:n_2 x=\frac{n_1\pi}{z}$
(ie if it doesn't work for two integers, the third won't add much. If it doesn't work for integers it sure as heck won't work for natural numbers, which are a subset.)

A: No, this is not true.
Let $x = \frac{\pi}{4}$, $x \ne \frac{\pi}{2}$.
Let $n_1 = 2$ and $n_2 = 1$.
Then,
$xn_1 = \frac{\pi}{4}\cdot 2 = \frac{\pi}{2} = \frac{\pi}{2}\cdot 1 = \frac{\pi}{2}n_2$
