EDIT: This solution is wrong! The error is highlighted below (thanks for Zaricuse and Rahul).
There is no solution.
Given $t$, define a function $$f(n) = \frac{\sin(nt)}{n}.$$
Its derivative is $$f'(n) = \frac{\cos(nt)nt - \sin(nt)}{n^2}.$$
If $f(n) = f(n+2)$ then the derivative must have a zero somewhere in between, say $f'(m) = 0$. That implies $$\tan(mt) = mt.$$ However, for $x \in (0,\pi)$ we have $\tan(x) > x$ (consider the Taylor expansion of $\tan$, for example). The unfortunate error is that although $t < \pi$, $x = mt$ is not necessarily less than $\pi$.