## Precalculus (6th Edition) Blitzer

Only symmetric with respect to the line $\theta=\frac{\pi}{2}$. See graph.
Step 1. We are given the polar equation $r=2sin(3\theta)$. To test the symmetry with respect to the polar axis, let $\theta\to -\theta$. We have $r=2sin(-3\theta)$ or $r=-2sin(3\theta)$. Thus, the equation is not symmetric with respect to the polar axis. Step 2. To test the symmetry with respect to the line $\theta=\frac{\pi}{2}$, let $r\to -r$ and $\theta\to -\theta$; we have $-r=2sin(-3\theta)$ or $r=2sin(3\theta)$. Thus, the equation is symmetric with respect to the line $\theta=\frac{\pi}{2}$. Step 3. To test the symmetry with respect to the pole, let $r\to -r$; we have $-r=2sin(3\theta)$ or $r=-2sin(3\theta)$. Thus, the equation is not symmetric with respect to the pole. Step 4. We can graph the equation as shown in the figure.