What is the Cartesian form of #r-5theta = sin^3theta+sectheta #?

To transform an expression in polar form to Cartesian (or rectangular) form, we need the formulas that relate #r# and #theta# to #x# and #y#.

Step 1. Recall the relationships between polar and Cartesian.

#x^2+y^2=r^2#

#x=rcos(theta)# or #cos(theta)=x/r# or #theta=cos^-1(x/r)#

#y=rsin(theta)# or #sin(theta)=y/r# or #theta=sin^-1(y/r)#

#tan(theta)=y/x#

These relationships can be determined from drawing a point in one of two ways, as radius and angle or as #x#-coordinate and #y#-coordinate as seen below.

Step 2. Convert each term from polar to Cartesian.
We are given #r-5theta=sin^3(theta)+sec(theta)#. Let's take each part and put it terms of #x# and #y#.

#r=sqrt(x^2+y^2)#

#5theta=5sin^-1(y/r)#

Plugging #r=sqrt(x^2+y^2)# into the previous gives

#5theta=5sin^-1(y/sqrt(x^2+y^2))#

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

#sec(theta)=1/cos(theta)=r/x=sqrt(x^2+y^2)/x#

Step 3. Plug these back into the original.

#r-5theta=sin^3(theta)+sec(theta)# becomes

#sqrt(x^2+y^2)-5sin^-1(y/sqrt(x^2+y^2))=(y/sqrt(x^2+y^2))^3+sqrt(x^2+y^2)/x#

You could spend a lot of time trying to simplify and trying to find a function #y# in terms of the independent variable #x#, but this is not what your question asked. So this is a great place to stop!