Answer
The tangent line is horizontal at the points $~~(\frac{3\sqrt{2}}{2},\frac{\pi}{4})~~$ and $~~(-\frac{3\sqrt{2}}{2},\frac{3\pi}{4})$
The tangent line is vertical at the points $~~(3,0)~~$ and $~~(0, \frac{\pi}{2})$
Work Step by Step
$r = 3~cos~\theta$
Note that this is the equation of a circle and a circle is completed as $\theta$ goes from $0 \leq \theta \lt \pi$. For other values of $\theta$, the points on the circle are repeated.
$x = r~cos~\theta = 3~cos^2~\theta$
$y = r~sin~\theta = 3~sin~\theta~ cos~\theta$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3cos^2~\theta-3sin^2~\theta}{-6~cos~\theta~sin~\theta} = \frac{sin^2~\theta-cos^2~\theta}{2~cos~\theta~sin~\theta}$
When the tangent line is horizontal, $\frac{dy}{dx} = 0$
We can find the values of $\theta$ when $\frac{dy}{dx} = 0$:
$\frac{dy}{dx} = \frac{sin^2~\theta-cos^2~\theta}{2~cos~\theta~sin~\theta} = 0$
$sin^2~\theta-cos^2~\theta = 0$
$sin^2~\theta = cos^2~\theta$
$tan^2~\theta = 1$
$tan~\theta = \pm 1$
$\theta = \frac{\pi}{4}, \frac{3\pi}{4}$
When $\theta = \frac{\pi}{4}$:
$r = 3~cos~\frac{\pi}{4} = \frac{3\sqrt{2}}{2}$
When $\theta = \frac{3\pi}{4}$:
$r = 3~cos~\frac{3\pi}{4} = -\frac{3\sqrt{2}}{2}$
The tangent line is horizontal at the points $~~(\frac{3\sqrt{2}}{2},\frac{\pi}{4})~~$ and $~~(-\frac{3\sqrt{2}}{2},\frac{3\pi}{4})$
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 $0$:
$2~cos~\theta~sin~\theta = 0$
$sin~\theta = 0~~~$ or $~~~cos~\theta = 0$
$\theta = 0, \frac{\pi}{2}$
When $\theta = 0$:
$r = 3~cos~0 = 3$
When $\theta = \frac{\pi}{2}$:
$r = 3~cos~\frac{\pi}{2} = 0$
The tangent line is vertical at the points $~~(3,0)~~$ and $~~(0, \frac{\pi}{2})$
