If $$x = 2\sin\theta - \sin 2\theta$$ and $$y = 2\cos\theta - \cos 2\theta$$, $$\theta \in [0, 2\pi]$$, then $$\frac{d^2y}{dx^2}$$ at $$\theta = \pi$$ is:

We are given the parametric equations $$x = 2\sin\theta - \sin 2\theta$$ and $$y = 2\cos\theta - \cos 2\theta$$, and we need to find $$\frac{d^2y}{dx^2}$$ at $$\theta = \pi$$.

First, we find the first derivatives with respect to $$\theta$$:

$$\frac{dx}{d\theta} = 2\cos\theta - 2\cos 2\theta$$

$$\frac{dy}{d\theta} = -2\sin\theta + 2\sin 2\theta$$

Now, $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-\sin\theta + \sin 2\theta}{\cos\theta - \cos 2\theta}$$

Using sum-to-product formulas:

Numerator: $$\sin 2\theta - \sin\theta = 2\cos\frac{3\theta}{2}\sin\frac{\theta}{2}$$

Denominator: $$\cos\theta - \cos 2\theta = 2\sin\frac{3\theta}{2}\sin\frac{\theta}{2}$$

Therefore: $$\frac{dy}{dx} = \frac{2\cos\frac{3\theta}{2}\sin\frac{\theta}{2}}{2\sin\frac{3\theta}{2}\sin\frac{\theta}{2}} = \cot\frac{3\theta}{2}$$

(The $$\sin\frac{\theta}{2}$$ terms cancel; at $$\theta = \pi$$, $$\sin\frac{\pi}{2} = 1 \neq 0$$, so this is valid.)

For the second derivative, we use: $$\frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta}}$$

$$\frac{d}{d\theta}\left(\cot\frac{3\theta}{2}\right) = -\csc^2\frac{3\theta}{2} \cdot \frac{3}{2} = -\frac{3}{2}\csc^2\frac{3\theta}{2}$$

At $$\theta = \pi$$:

$$\csc^2\frac{3\pi}{2} = \frac{1}{\sin^2\frac{3\pi}{2}} = \frac{1}{(-1)^2} = 1$$

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

Also, $$\frac{dx}{d\theta}\bigg|_{\theta=\pi} = 2\cos\pi - 2\cos 2\pi = 2(-1) - 2(1) = -4$$

Therefore: $$\frac{d^2y}{dx^2}\bigg|_{\theta=\pi} = \frac{-\frac{3}{2}}{-4} = \frac{3}{8}$$

The answer is Option A: $$\frac{3}{8}$$.