I use both #r and theta >= 0# only.
So, #arctan (y/x) in [ 0, pi ], sans pi/2#
#r = theta - 4 sin theta + cos^2theta = theta# + a periodic function
of period #2pi#. So, r is non-periodic.
Convert using
#( x, y ) = r ( cos theta, sin theta ). 0 <= r = sqrt ( x^2 + y^2 ) and#
#0 <= theta = kpi + arctan ( y/x ), l = 0, 1, 2, 3, ...#.
#sqrt ( x^2 + y^2 ) = kpi + arctan ( y/x )#
#- 4 x/sqrt ( x^2 + y^2 ) + y^2/(x^2 + y^2 )#, giving
#0 <= (tan)^(-1) ( y/x ) = kpi + arctan ( y/x ) #
#= 4 x/sqrt ( x^2 + y^2 ) - y^2/(x^2 + y^2 )#
#+ sqrt ( x^2 + y^2 )#, k = 0, 1, 2, 3, ..#, piecewise, for successive
half rotations, about O.
Graph, for k = 0 using conventional #arctan ( y/x ) in ( - pi/2, pi/2 )# )#.graph{arctan ( y/x )-4 x/sqrt ( x^2 + y^2 ) + y^2/(x^2 + y^2 )=0[-1000 1000 -500 500]}
Graph, for the assumed #theta to theta + pi/2# non- negative
#arctan ( y/x) in ( 0, pi )#, sans #pi/2 # .
graph{pi/2+arctan ( y/x )+4y/sqrt ( x^2 + y^2 ) + x^2/(x^2 + y^2 )=0[-1000 1000 -500 500]}
Observe that this is confined to #Q_3 and Q_4#, whereas, the first
that was expected to be in #Q_1 and Q_4# is not so.
