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?
 A: You have calculated
$$
\frac{dx}{d\theta}=-\sin\theta(1+4\cos\theta)
$$
and
$$
\frac{dy}{d\theta}=\cos\theta+2\cos2\theta.
$$
Since $\cos2\theta=2\cos^2\theta-1$, the $\frac{dy}{d\theta}$ simplifies to
$$
\frac{dy}{d\theta}=\cos\theta+4\cos^2\theta-2.
$$
Vertical and horizontal tangents
In fact, we do not need to find the values of $\theta$ if we want to find the points where we have horizontal/vertical tangents. It is enough to know $\cos\theta$ and $\sin\theta$ and then use the formulas
$$
x=r\cos\theta=(1+2\cos\theta)\cos\theta,\quad\text{and}\quad y=r\sin\theta=(1+2\cos\theta)\sin\theta.
$$
Anyways, I indicate how to solve the equations. First, we can assume that $0\leq \theta<2\pi$, since $r=1+2\cos\theta$ is $2\pi$ periodic.
Now, one should solve $\frac{dx}{d\theta}=0$ to find the vertical tangent lines. For the vertical ones, $dx/d\theta=0$. Since the product $\sin\theta(1+4\cos\theta)=0$ if and only if (at least) one of its factors is zero, we find that either $\sin\theta=0$ (this gives $\theta=0$ and $\theta=\pi$), or $\cos\theta=-1/4$ (this gives $\theta=\arccos(-1/4)$ or $\theta=2\pi-\arccos(-1/4)$. In total, we get four values of $\theta$.
To find the horizontal tangent lines we solve $dy/d\theta=0$. We note that $dy/d\theta$ is a second degree polynomial in $\cos\theta$. Thus, with $t=\cos\theta$, we should solve $4t^2+t-2=0$. This has solutions $t=\frac{1}{8}(-1\pm\sqrt{33})$. So, we find $\theta$ by solving
$$
\cos\theta=\frac{1}{8}(-1+\sqrt{33})\approx 0.59,\quad\text{and},\quad \cos\theta=\frac{1}{8}(-1-\sqrt{33})\approx -0.84.
$$
To calculate $\sin\theta$ you can use the fact that $\cos^2\theta+\sin^2\theta=1$. I leave it to you to fill in the rest of the details. If you succeed, you will probably find something like the picture below, where I have drawn the curve $r=1+2\cos\theta$ together with the points I got by doing these calculations (red for horizontal tangents, green for vertical ones).

A: 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 $\theta$.
