Maths(Polar Coordinates - maximum value)

Find the maximum value of the y-corrdinate of points on the limacon r = 1 + 2cos(theta).
sub x=rcos(theta)
y=rsin(theta)
The equation becomes
x^2+y^2=1+2x/ √(x^2+y^2)
x^2+y^2=√(x^2+y^2)+2x
2x+(2y)(dy/dx)=x/√(x^2+y^2)+2
Let dy/dx=0
x/√(x^2+y^2)+2=2x
or cosθ+2=2rcosθ
cosθ=2/(2r-1)
sub into r = 1 + 2cosθ
r=1+4/(2r-1)
2r^2-r-2r+1-4=0
2r^2-3r-3=0
r=2.18614
θ=53.6248
y=rsinθ=1.7602