Answer
The slope of the tangent line to the given polar curve at that point is equal to: $$\frac{\sqrt 3}{9}$$
Work Step by Step
1. Write the formula for the slope of the tangent line: $$\frac{dy}{dx} = \frac{\frac{dr}{d\theta}sin(\theta) + rcos(\theta)}{\frac{dr}{d\theta}cos(\theta) - rsin(\theta)} $$
2. Calculate $\frac{dr}{d\theta}$ and substitute the equation into the formula:
$\frac{dr}{d\theta} = \frac{d(1 + 2cos(\theta))}{d\theta} = -2sin(\theta)$
$\frac{dy}{dx} = \frac{-2sin(\theta)sin(\theta) + (1 + 2cos(\theta))cos(\theta)}{-2sin(\theta)cos(\theta) - (1 + 2cos(\theta))sin(\theta)} = \frac{-2sin^2(\theta) + cos(\theta) + 2cos^2(\theta)}{-2sin(\theta)cos(\theta) - sin(\theta) -2cos(\theta)sin(\theta)}$
$\frac{dy}{dx} = \frac{2cos(2\theta) + cos(\theta)}{-2sin(2\theta) - sin(\theta)}$
3. Calculate the slope when $\theta = \pi/3$
$\frac{dy}{dx} = \frac{2cos(2\frac{\pi}{3}) + cos(\frac{\pi}{3})}{-2sin(2\frac{\pi}{3}) - sin(\frac{\pi}{3})} = \frac{2(-1/2) + 1/2}{-2(\frac{\sqrt 3}{2}) - \frac{\sqrt 3}{2}}$
$\frac{dy}{dx} = \frac{-1/2}{-\sqrt 3 - \frac{\sqrt 3}{2}} = \frac{-1/2}{-\frac{3\sqrt 3}{2}} = \frac{1}{3\sqrt 3} \times \frac{\sqrt 3}{\sqrt 3} = \frac{\sqrt 3}{3\sqrt 3 \sqrt 3} = \frac{\sqrt 3}{9}$
