Answer
The tangent line is horizontal at the points $~~(\frac{3}{2},\frac{\pi}{3}),~~$ $(\frac{3}{2},\frac{5\pi}{3}),~~$ and $~~(0,\pi)$
The tangent line is vertical at the points $~~(2,0),~~$ $(\frac{1}{2}, \frac{2\pi}{3}),~~$ and $~~(\frac{1}{2}, \frac{4\pi}{3})$
Work Step by Step
$r = 1+~cos~\theta$
Note that $r \geq 0$ for all values of $\theta$. Thus we need to consider values of $\theta$ from $0 \leq \theta \lt 2\pi$. For other values of $\theta$, the points on the curve are repeated.
$x = r~cos~\theta = cos~\theta+~cos^2~\theta$
$y = r~sin~\theta = sin~\theta+sin~\theta~ cos~\theta$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{cos~\theta+cos^2~\theta-sin^2~\theta}{-sin~\theta-2~cos~\theta~sin~\theta}$
When the tangent line is horizontal, $\frac{dy}{dx} = 0$
We can find the values of $\theta$ when $\frac{dy}{dx} = 0$:
$\frac{dy}{dx} = \frac{cos~\theta+cos^2~\theta-sin^2~\theta}{-sin~\theta-2~cos~\theta~sin~\theta} = 0$
$cos~\theta+cos^2~\theta-sin^2~\theta = 0$
$cos~\theta+cos^2~\theta-(1-cos^2~\theta) = 0$
$2~cos^2~\theta+cos~\theta-1 = 0$
$(2~cos~\theta-1)(cos~\theta+1) = 0$
$cos~\theta = \frac{1}{2}~~~$ or $~~~cos~\theta = -1$
$\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \pi$
When $\theta = \frac{\pi}{3}$:
$r = 1+cos~\frac{\pi}{3} = \frac{3}{2}$
When $\theta = \frac{5\pi}{3}$:
$r = 1+cos~\frac{5\pi}{3} = \frac{3}{2}$
When $\theta = \pi$:
$r = 1+cos~\pi = 0$
The tangent line is horizontal at the points $~~(\frac{3}{2},\frac{\pi}{3}),~~$ $(\frac{3}{2},\frac{5\pi}{3}),~~$ and $~~(0,\pi)$
When the tangent line is vertical, the denominator of $\frac{dy}{dx}$ is $0$
We can find the values of $\theta$ when the denominator of $\frac{dy}{dx}$ is $0$:
$-sin~\theta-2~cos~\theta~sin~\theta = 0$
If $sin~\theta \neq 0$:
$2~cos~\theta~sin~\theta = -sin~\theta$
$cos~\theta = -\frac{1}{2}$
$\theta = \frac{2\pi}{3}, \frac{4\pi}{3}$
If $sin~\theta = 0,~~$ then $\theta = 0, \pi$
However, above we stated that the tangent line is horizontal when $\theta = \pi$ We can use L'Hospital's Rule to verify the following limit:
$\lim\limits_{\theta \to \pi}\frac{dy}{dx}$
$= \lim\limits_{\theta \to \pi}\frac{cos~\theta+cos^2~\theta-sin^2~\theta}{-sin~\theta-2~cos~\theta~sin~\theta}$
$= \lim\limits_{\theta \to \pi}\frac{-sin~\theta-2~cos~\theta~sin~\theta-2~sin~\theta~cos~\theta}{-cos~\theta+2~sin^2~\theta-2~cos^2~\theta}$
$= \frac{0}{-1}$
$= 0$
Both the numerator and denominator approach $0$ as $\theta \to \pi$, but since $\lim\limits_{\theta \to \pi}\frac{dy}{dx} = 0$, the tangent line is horizontal at $\theta = \pi$
When $\theta = 0$:
$r = 1+cos~0 = 2$
When $\theta = \frac{2\pi}{3}$:
$r = 1+cos~\frac{2\pi}{3} = \frac{1}{2}$
When $\theta = \frac{4\pi}{3}$:
$r = 1+cos~\frac{4\pi}{3} = \frac{1}{2}$
The tangent line is vertical at the points $~~(2,0),~~$ $(\frac{1}{2}, \frac{2\pi}{3}),~~$ and $~~(\frac{1}{2}, \frac{4\pi}{3})$
