Answer
Only 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 $r =3cos(-\theta)$ or $r =3cos(\theta)$. 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 $-r =3cos(-\theta)$ or $r =-3cos(\theta)$. Thus the equation is not necessarily 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 $-r =3cos(\theta)$ or $r =-3cos(\theta)$. Thus the equation is not necessarily symmetric with respect to the pole.
Step 4. Use test points with $0\leq\theta\leq\pi$, we can graph the function as shown in the figure.
