Answer
$(-\infty,\infty)$
Work Step by Step
$r(\theta) = (\theta - 2\sin\theta)\textbf i + (1-2\cos \theta)\textbf j \\
\Rightarrow r'(\theta) = (1-2\cos \theta)\textbf i + 2\sin \theta \textbf j$
To get $r'(\theta) = 0\textbf i + 0 \textbf j$ we need to find value of $\theta$ for which $(1-2\cos \theta) =0$ and $2\sin \theta = 0$ simultaneously.
As there is no such value of $\theta$ possible, so, the curve is smooth on the interval $(-\infty,\infty).$
