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This is essentially asking you to find where the derivative of r is horizontal AKA 0.
\[r(\theta)'=(2 + \cos (\theta))' = -\sin(\theta)\]
The translation by +2 simply moves the graph up and down so it doesn't affect the tangent. We are left with a negative (upside-down) sine graph. So where is...
\[-\sin(\theta)=0??\]
Let's look at a graph. Notice in the pic attachment the curve touches down to zero in increments of pi, forever.
A generic way of saying this is that the graph of 2 + cos(theta) has horizontal tangents at k*pi where k is any integer (-1, 0, 1, 2, etc)