I need some help finding the slope of the tangent line of a graph!

I really need some help with this problem. I've tried several different methods to get the answer but I can't get the answer!

Find the slope of the tangent line to the graph of r = 3 + 3 cos θ at the point on the graph where θ = (1/6)π.

I'm assuming that r = 3 + 3cos(θ) is supposed to be polar, so if that's not true, then this answer will be wrong. Anyway...

As I'm sure you know, the formula for the slope of a tangent line to a graph at a given point is just the derivative of the function evaluated at that point. So, we need to know the derivative of r = 3 + 3cos(θ). For polar coordinates, we have the following formula for a derivative:

dy/dx = (dy/dθ)/(dx/dθ)

we also need to know how to convert from polar to rectangular:
x(θ) = rcos(θ)
y(θ) = rsin(θ)

since we know r = 3 + 3cos(θ), we have:
x(θ) = rcos(θ) = (3 + 3cos(θ))cos(θ) = 3cos(θ) + 3cos²(θ)
y(θ) = rsin(θ) = (3 + 3cos(θ))sin(θ) = 3sin(θ) + 3sin(θ)cos(θ) = 3(2sin(θ) + sin(2θ)) / 2

thus, we have:
x'(θ) = -6sin(θ)cos(θ) - 3sin(θ) = -3(sin(θ) + sin(2θ))
y'(θ) = 3(cos(θ) + cos(2θ))

(note that I used the trig identity sin(θ)cos(θ) = sin(2θ)/2 a couple of times)

Now we have what we need to take the derivative:
dy/dx = 3(cos(θ) + cos(2θ)) / -3(sin(θ) + sin(2θ)) = -(cos(θ) + cos(2θ)) / (sin(θ) + sin(2θ))

So to finish the problem, just plug in θ = π/6:
-(cos(π/6) + cos(2π/6)) / (sin(π/6) + sin(2π/6))
-(sqrt(3)/2 + 1/2) / (1/2 + sqrt(3)/2)
-1

So, the slope of the tangent line is -1

Hope this helps!