<s>[INST] Where does $r = 1 + 2\cos(\theta)$ have tangents? Where does:
$$r = 1 + 2\cos(\theta)$$
Have horizontal and vertical tangent lines?
$x = r\cos(\theta) = \cos(\theta) + 2\cos^2(\theta)$
$y = r\sin(\theta) = \sin(\theta) + 2\sin(\theta)\cos(\theta)$
$$dx/d\theta = -\sin(\theta) - 4\sin(\theta)\cos(\theta)$$
$$dy/d\theta = \cos(\theta) + 2\cos(2\theta)$$
By hand it is very hard to solve  $dy/d\theta = 0, dx/d\theta = 0$
How can it be done?
 [/INST] Note that $\cos(2\theta) = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1$ so
$$\frac{dy}{d\theta} = 0 \Leftrightarrow \cos\theta + 4\cos^2 \theta - 2 = 0$$
Now compute the roots of $4x^2+x-2$ to get equations for $\cos \theta$ and apply $\arccos$ to find $\theta$.
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