well, hmmm to check for polar symmetry, hmmm say we simply replace the value for either "r" or "θ" or both in the original equation, if the equation rendered is the same as the original, there you have it, symmetry, so let's do that.
[tex]\underline{\textit{testing for symmetry to the }\frac{\pi }{2}~line\qquad \qquad r=-r~~,~~\theta =-\theta } \\\\\\ r=-7+2cos(3\theta )\implies (-r)=-7+2cos[3(-\theta )]\implies -r=-7+2cos(-3\theta ) \\\\\\ \stackrel{symmetry~identity}{-r=-7+2\stackrel{\downarrow }{cos(3\theta )}}\implies r=7-2cos(3\theta )\impliedby \textit{woopsie, no dice} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\underline{\textit{testing for symmetry to the polar axis}\qquad \qquad \theta =-\theta} \\\\\\ r=-7+2cos(3\theta )\implies r=-7+2cos[3(-\theta )]\implies r=-7+2cos(-3\theta ) \\\\\\ \stackrel{symmetry~identity}{r=-7+2\stackrel{\downarrow }{cos(3\theta )}}\impliedby \textit{low and behold\qquad \Large \checkmark}[/tex]
now, we could run a test for "the pole" or namely the origin by simply changin r = -r, however we already know where the symmetry is, so no need.
Check the picture below.
