i have no idea on how to go about solving this, please help!
Let's call (theta) = t for convenience.
sin(3t) = sin(2t + t) = sin(2t)cos(t) + sin(t)cos(2t)
= 2sin(t)cos(t)*cos(t) + sin(t)*(1 - 2sin^2(t))
= 2sin(t)cos^2(t) + sin(t) - 2sin^3(t)
= 2sin(t)(1 - sin^2(t)) + sin(t) - 2sin^3(t)
= 2sin(t) - 2sin^3(t) + sin(t) - 2sin^3(t)
= 3sin(t) - 4sin^3(t)
So your polar equation is:
r = 6sin(t) - 8sin^3(t)
Now,
r = (x^2 + y^2)^{1/2}
t = atn(y/x)
This is probably not the easiest way to go. I mean, sin(atn(y/x))? It can be done, but we have to do this for four different quadrants. Let's try this:
y = r*sin(t)
sin(t) = y/r = y/(x^2 + y^2)^{1/2}
So...
(x^2 + y^2)^{1/2} = 6*y/(x^2 + y^2)^{1/2} - 8*y^3/(x^2 + y^2)^{3/2}
I'm going to clear the fractions and multiply both sides by (x^2 + y^2)^{3/2}:
(x^2 + y^2)^2 = 6y(x^2 + y^2) - 8y^3
And you can simplify from here as you like. It's not going to simplify into anything pretty anyway.
-Dan