cos 3θ = sin 3θ since both = r 

 Divide both sides by cos 3θ



 
1 = tan(3θ)

So 3θ has referent angle 45° or , The tangent is positive in 
QI and QIII 

Since we are requiring that  0 ≤ θ ≤ π, that means that
                             0 ≤ 3θ ≤ 3π

So 3θ can be , , and 

So θ can be , , and , which reduces to  

We find the value of r for each of those three values for θ, by substituting
in either equation for r

when θ = ,r = cos 3θ = cos = cos = 

So the first point is (r,θ) =  

when θ = ,r = cos 3θ = cos = cos = 

So the second point is (r,θ) = 

when θ = ,r = cos 3θ = cos = 

So the third point is (r,θ) = 

Edwin