$|\frac{\sin(nx)}{n\sin(x)}|\le1\forall x\in\mathbb{R}-\{\pi k: k\in\mathbb{Z}\}$ Find all real number $n$ such that  $|\frac{\sin(nx)}{n\sin(x)}|\le1\forall x\in\mathbb{R}-\{\pi k: k\in\mathbb{Z}\}$.

The inequality is true for all $n \in \mathbb{Z}^+$, which also implies immediately that it is true for $n \in \mathbb{Z}^-$. The proof follows immediately from induction.
For $n = 1$, it is true.
Assume it is true for $n=k$ i.e. we have $$\left\lvert \dfrac{\sin(kx)}{k \sin(x)} \right\rvert \leq 1$$
Then $$\sin((k+1)x) = \sin(kx) \cos(x) + \cos(kx) \sin(x)$$
$$\left \vert \sin((k+1)x) \right \vert = \left \vert \sin(kx) \cos(x) + \cos(kx) \sin(x) \right \vert \leq \left \vert \sin(kx) \cos(x) \right \vert + \left \vert \cos(kx) \sin(x) \right \vert\\ \leq k \left \vert \sin(x) \right \rvert + \left \vert \sin(x) \right \rvert$$
If $n \notin \mathbb{Z}^+$, then $\left\lvert \dfrac{\sin(kx)}{k \sin(x)} \right\rvert $ can be made arbitrarily large by choosing $x$ close to say $\pi$.
