Find the values of θ at which there are horizontal tangent lines on the graph of r = 1 + sin θ

Question

a) π/2, 3π/2, π/6, 5π/6
b) π/2, 7π/6, 11π/6
c) π/2, 3π/2, 7π/6, 11π/6
d) 3π/2, 2π/3, 4π/3

Answer 1

horizontal lines have a slope of zero, right?
y = r sinθ so dy/dθ = r' sinθ + r cosθ
x = r cosθ so dx/dθ = r' cosθ - r sinθ

dy/dx = (dy/dθ) / (dx/dθ)
dy/dθ = (cosθ)(sinθ) + (1+sinθ)cosθ
= cosθ(1+2sinθ)
dx/dθ = (cosθ)(cosθ) - (1+sinθ)(sinθ)
= cos^2θ - sin^2θ - sinθ = cos2θ - sinθ
dy/dx = 0 where dy/dθ = 0 and dx/dθ ≠ 0

dy/dθ = 0 when
cosθ = 0 (θ = π/2, 3π/2)
1+2sinθ = 0 (θ = 7π/6, 11π/6)

dx/dθ = 0 when θ = 3π/2

so y'=0 at π/2, 7π/6, 11π/6

you can verify this from the graph at

https://www.wolframalpha.com/input/?i=r%3D1%2Bsin%CE%B8