1 − cos θ = 1 + sin θ  , since both = r
    −cos θ = sin θ
Divide both sides by cos θ
    = 
   -1 = tan(θ)

   
So θ has two possible values where 0 ≤ θ ≤ 2π. the tangent is negative
in Q2 and Q4 and the reference angle is 45° or , so θ is 
in Q2 or  in Q4.
 
So the smaller value of θ is , and the value of r is gotten
by substituting  for θ in either of the original equations:
r = 1 − cos θ = 1 - cos = 1 -  = 1 +  = 
So the first point is (r,θ) = (, )
The larger value of θ is , and the value of r is gotten
by substituting  for θ in either of the original equations:
r = 1 − cos θ = 1 - cos = 1 -  = 1 -  = 
So the second point is (r,θ) = (, )

Edwin