Precalculus (6th Edition) Blitzer

The graph does not necessarily have symmetry with respect to the polar axis, the line $\theta =\frac{\pi }{2}$, or the pole. See the graph below:
We look for symmetry by making the following substitutions: (a) $\theta \to - \theta$ :$$r=1-3\sin (- \theta ) \quad \Rightarrow \quad r=1+3\sin \theta$$Thus, the graph does not necessarily have symmetry with respect to the polar axis. (b) $r \to -r, \quad \theta \to -\theta$ :$$-r =1-3\sin (-\theta ) \quad \Rightarrow \quad r=-1-3\sin \theta$$Thus, the graph does not necessarily have symmetry with respect to the line $\theta=\frac{\pi}{2}$. (c) $r \to -r$ :$$-r =1-3\sin \theta \quad \Rightarrow \quad r=-1+3\sin \theta$$Thus, the graph does not necessarily have symmetry with respect to the pole.