Convert the polar equation to rectangular coordinates. (use variables x and y as needed.)

r = 3 − cos θ

[tex]cos(\theta) = x/\sqrt{x^2+y^2} = x/r, and r = \sqrt{ x^2+y^2} so

r = 3 - x/r ==> r^2 + 3r + x = 0,
[\tex]
r = solutions of quadratic and finally
[tex]
r = \sqrt{ x^2+y^2} = solutions,
[\tex]

you could eventually write y(x)! Not sure you're still looking at this question though! ?:)