Answer
Equation may not have symmetry; see graph.
Work Step by Step
Step 1. To test the symmetry with respect to the polar axis, replace $(r,\theta)$ with $(r,-\theta)$; we have $r =2+2sin(-\theta)$ or $r =2-2sin(\theta)$ . Thus the equation may not be symmetric with respect to the polar axis.
Step 2. To test the symmetry with respect to the line $\theta=\frac{\pi}{2}$, replace $(r,\theta)$ with $(-r,-\theta)$; we have $-r =2+2sin(-\theta)$ or $r =-2+2sin(\theta)$ . Thus the equation may not be symmetric with respect to the line $\theta=\frac{\pi}{2}$.
Step 3. To test the symmetry with respect to the pole, replace $(r,\theta)$ with $(-r,\theta)$, we have $-r =2+2sin(\theta)$ or $r =-2-2sin(\theta)$ . Thus the equation may not be symmetric with respect to the pole.
Step 4. Use test points with $0\leq\theta\leq2\pi$, we can graph the equation as shown in the figure.
