What is the Cartesian form of #r^2-rtheta = 2costheta+3tantheta #?

Like #r, theta >= 0#. This is important, for reading this answer,

wherein arctan values #in [ 0, pi ]# and not #[ -pi/2, pi/2 ]#.

And, you can choose any #theta# in any Q.

The chosen #theta = kpi + arctan( y/x), arctan ( y/x ) in [ 0, pi ]#,

for a specific k, from { 0, 1, 2, 3, ...}

Use

#( x, y ) = r ( cos theta, sin theta ), r = sqrt ( x^2 +y^2 ) >=0# and

#theta = kpi + arctan (y/x), theta in Q_1, Q_2 or Q_4#, choosing

befitting k from 0, 1, 2, 3, .... Now,

#r^2 - rtheta = 2 cos theta + 3 tan theta# converts to k-specific

#(x^2 + y^2 )^1.5 - ( x^2 + y^2 )( kpi + arctan ( y/x ))#

#= 2 x + 3 (y/x) sqrt( x^2 + y^2 )#

See graph, with k = 0.
graph{(x^2 + y^2 )^1.5 - ( x^2 + y^2 )( arctan ( y/x )) - 2 x - 3(y/x)sqrt( x^2 + y^2 )=0}

Note that the graph is non-periodic, despite that the joint period of

#cos theta and tan theta# is #2pi#. .

Graph for #theta to theta + 2pi#, in the next round,

graph{(x^2 + y^2 )^1.5 - ( x^2 + y^2 )( 2pi + arctan ( y/x )) - 2 x - 3(y/x)sqrt( x^2 + y^2 )=0}

Graph for #theta to theta + 4pi#
graph{(x^2 + y^2 )^1.5 - ( x^2 + y^2 )( 4pi + arctan ( y/x )) - 2 x - 3(y/x)sqrt( x^2 + y^2 )=0}

Slide the graphs #uarr larr darr rarr# for graphs, in grandeur.