Answer
Equation may only be symmetric with respect to the polar axis.
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 $r =1-2cos(-\theta)$ or $r =1-2cos(\theta)$ . Thus the equation is 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 $-r =1-2cos(-\theta)$ or $r =-1+2cos(\theta)$ . 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 $-r =1-2cos(\theta)$ or $r =-1+2cos(\theta)$ . Thus the equation may not be symmetric with respect to the pole.
Step 4. Using test points with $0\leq\theta\leq2\pi$, we can graph the equation as shown in the figure.
