Answer
Vertical Tangent Lines at: $(0,\frac{\pi}{2})$, $(3,0)$.
Horizontal Tangent Lines at: $(\frac{-3\pi}{2},\frac{3\pi}{4})$, $(\frac{3\pi}{2},\frac{\pi}{4})$.
Work Step by Step
Given:
$r=3cos(\theta)$
Use the equation for tangent line slope for polar coordinates:
$TLS=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta}{\frac{dr}{d\theta}cos\theta-rsin\theta}$
Find $\frac{dr}{d\theta}$:
$\frac{dr}{d\theta}=(3cos(\theta))^{\prime}={-3sin(\theta)}$
Plug in for $r$ and $\frac{dr}{d\theta}$:
$TLS\vert_{\theta}=\frac{(-3sin(\theta))sin(\theta)+(3cos(\theta))cos(\theta)}{({-3sin(\theta)})cos(\theta)-(3cos(\theta))sin(\theta)}=\frac{(-3sin(\theta))sin(\theta)+(3cos(\theta))cos(\theta)}{({-3sin(\theta)})cos(\theta)-(3cos(\theta))sin(\theta)}$
$TLS\vert_{\theta}=\frac{-sin^2\theta+cos^2\theta}{{-2sin{\theta}}cos\theta}=\frac{-sin^2\theta+cos^2\theta}{{-2sin(2\theta)}}$
Find the horizontal tangent lines (when the denominator = 0):
$-sin^2\theta+cos^2\theta=0$
-->$sin^2\theta=cos^2\theta$
-->$sin\theta=cos\theta$
This is true when $\theta=\frac{\pi}{4}$ and $\frac{3\pi}{4}$
Now find the r-values for these points and form the ordered pairs in $(r,\theta)$:
$r=3cos(\frac{\pi}{4})=\frac{3\pi}{2}$
$r=3cos(\frac{3\pi}{4})=\frac{-3\pi}{2}$
Horizontal Tangent Lines at: $(\frac{-3\pi}{2},\frac{3\pi}{4})$, $(\frac{3\pi}{2},\frac{\pi}{4})$.
Now find the vertical tangent lines (when the numerator = 0):
${{-2sin(2\theta)}}=0$
-->$sin(2\theta)=0$
This is true when $\theta=0$ and $\frac{\pi}{2}$
Now find the r-values for these points and form the ordered pairs in $(r,\theta)$:
$r=3cos(0)=3$
$r=3cos(\frac{\pi}{2})=0$
Vertical Tangent Lines at: $(0,\frac{\pi}{2})$, $(3,0)$.