Answer
The slope of the tangent line is $\frac{1}{\sqrt{3}}$
Work Step by Step
$r = 2~cos~\theta$
$x = r~cos~\theta = 2~cos^2~\theta$
$y = r~sin~\theta = 2~sin~\theta~ cos~\theta$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2cos^2~\theta-2sin^2~\theta}{-4~cos~\theta~sin~\theta} = \frac{sin^2~\theta-cos^2~\theta}{2~cos~\theta~sin~\theta}$
When $\theta = \frac{\pi}{3}$:
$\frac{dy}{dx} = \frac{sin^2~\theta-cos^2~\theta}{2~cos~\theta~sin~\theta}$
$\frac{dy}{dx} = \frac{sin^2~\frac{\pi}{3}-cos^2~\frac{\pi}{3}}{2~cos~\frac{\pi}{3}~sin~\frac{\pi}{3}}$
$\frac{dy}{dx} = \frac{(\frac{\sqrt{3}}{2})^2-(\frac{1}{2})^2}{2~(\frac{1}{2})~(\frac{\sqrt{3}}{2})}$
$\frac{dy}{dx} = \frac{\frac{3}{4}-\frac{1}{4}}{\frac{\sqrt{3}}{2}}$
$\frac{dy}{dx} = \frac{1}{\sqrt{3}}$
The slope of the tangent line is $\frac{1}{\sqrt{3}}$
