Given parametric equations x = 1 + sin(2θ) and y = 1 + cosθ + cos(2θ), find dy/dx at θ = π/3.

Given $x = 1 + \sin(2\theta)$.
Differentiate $x$ with respect to $\theta$:

$$\frac{dx}{d\theta} = \frac{d}{d\theta}(1 + \sin(2\theta))$$ $$\frac{dx}{d\theta} = 0 + \cos(2\theta) \cdot \frac{d}{d\theta}(2\theta)$$ $$\frac{dx}{d\theta} = 2\cos(2\theta)$$

Given $y = 1 + \cos\theta + \cos(2\theta)$.
Differentiate $y$ with respect to $\theta$:

$$\frac{dy}{d\theta} = \frac{d}{d\theta}(1 + \cos\theta + \cos(2\theta))$$ $$\frac{dy}{d\theta} = 0 - \sin\theta - \sin(2\theta) \cdot \frac{d}{d\theta}(2\theta)$$ $$\frac{dy}{d\theta} = -\sin\theta - 2\sin(2\theta)$$

Now, substitute the expressions for $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ into the formula for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{-\sin\theta - 2\sin(2\theta)}{2\cos(2\theta)}$$

Substitute $\theta = \frac{\pi}{3}$ into the expression for $\frac{dy}{dx}$:

$$\frac{dy}{dx}\Big|_{\theta=\frac{\pi}{3}} = \frac{-\sin(\frac{\pi}{3}) - 2\sin(2 \cdot \frac{\pi}{3})}{2\cos(2 \cdot \frac{\pi}{3})}$$

Recall the values:
$\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$
$\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$
$\cos(\frac{2\pi}{3}) = -\frac{1}{2}$

Substitute these values:

$$\frac{dy}{dx}\Big|_{\theta=\frac{\pi}{3}} = \frac{-\frac{\sqrt{3}}{2} - 2\left(\frac{\sqrt{3}}{2}\right)}{2\left(-\frac{1}{2}\right)}$$ $$\frac{dy}{dx}\Big|_{\theta=\frac{\pi}{3}} = \frac{-\frac{\sqrt{3}}{2} - \sqrt{3}}{-1}$$ $$\frac{dy}{dx}\Big|_{\theta=\frac{\pi}{3}} = \frac{-\frac{\sqrt{3}}{2} - \frac{2\sqrt{3}}{2}}{-1}$$ $$\frac{dy}{dx}\Big|_{\theta=\frac{\pi}{3}} = \frac{-\frac{3\sqrt{3}}{2}}{-1}$$ $$\frac{dy}{dx}\Big|_{\theta=\frac{\pi}{3}} = \frac{3\sqrt{3}}{2}$$

Comparing our derived expression for $\frac{dy}{dx}$ with the given options, Option A matches:

$$\text{Option A: } \frac{-\sin\theta - 2\sin(2\theta)}{2\cos(2\theta)} \text{ evaluated at } \pi/3$$

The final answer is $\text{Option A}$.