The equation of a normal to the curve, $$\sin y = x\sin\left(\frac{\pi}{3} + y\right)$$ at $$x = 0$$, is:

To find the equation of the normal to the curve $$\sin y = x \sin\left(\frac{\pi}{3} + y\right)$$ at $$x = 0$$, we first need to determine the point(s) on the curve where $$x = 0$$. Substituting $$x = 0$$ into the equation gives:

$$\sin y = 0 \cdot \sin\left(\frac{\pi}{3} + y\right) \implies \sin y = 0$$

This implies $$y = n\pi$$, where $$n$$ is an integer. So the points are $$(0, n\pi)$$ for any integer $$n$$. Common points include $$(0, 0)$$, $$(0, \pi)$$, $$(0, 2\pi)$$, etc.

Next, we need the slope of the tangent to the curve at these points. To find $$\frac{dy}{dx}$$, we differentiate both sides of the curve equation implicitly with respect to $$x$$. The equation is:

$$\sin y = x \sin\left(\frac{\pi}{3} + y\right)$$

Differentiating the left side:

$$\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$$

Differentiating the right side (using the product rule):

$$\frac{d}{dx}\left[x \sin\left(\frac{\pi}{3} + y\right)\right] = \left(1\right) \cdot \sin\left(\frac{\pi}{3} + y\right) + x \cdot \frac{d}{dx}\left[\sin\left(\frac{\pi}{3} + y\right)\right]$$

The derivative of $$\sin\left(\frac{\pi}{3} + y\right)$$ is:

$$\frac{d}{dx}\left[\sin\left(\frac{\pi}{3} + y\right)\right] = \cos\left(\frac{\pi}{3} + y\right) \cdot \frac{d}{dx}\left(\frac{\pi}{3} + y\right) = \cos\left(\frac{\pi}{3} + y\right) \cdot \frac{dy}{dx}$$

So the right side becomes:

$$\sin\left(\frac{\pi}{3} + y\right) + x \cos\left(\frac{\pi}{3} + y\right) \frac{dy}{dx}$$

Equating both sides:

$$\cos y \cdot \frac{dy}{dx} = \sin\left(\frac{\pi}{3} + y\right) + x \cos\left(\frac{\pi}{3} + y\right) \frac{dy}{dx}$$

Rearranging terms to isolate $$\frac{dy}{dx}$$:

$$\cos y \cdot \frac{dy}{dx} - x \cos\left(\frac{\pi}{3} + y\right) \frac{dy}{dx} = \sin\left(\frac{\pi}{3} + y\right)$$

$$\frac{dy}{dx} \left( \cos y - x \cos\left(\frac{\pi}{3} + y\right) \right) = \sin\left(\frac{\pi}{3} + y\right)$$

Solving for $$\frac{dy}{dx}$$:

$$\frac{dy}{dx} = \frac{\sin\left(\frac{\pi}{3} + y\right)}{\cos y - x \cos\left(\frac{\pi}{3} + y\right)}$$

This is the slope of the tangent at any point $$(x, y)$$ on the curve. Now, evaluate at $$x = 0$$ and $$y = n\pi$$.

First, at $$(0, 0)$$:

$$\frac{dy}{dx} = \frac{\sin\left(\frac{\pi}{3} + 0\right)}{\cos 0 - 0 \cdot \cos\left(\frac{\pi}{3} + 0\right)} = \frac{\sin\left(\frac{\pi}{3}\right)}{\cos 0} = \frac{\frac{\sqrt{3}}{2}}{1} = \frac{\sqrt{3}}{2}$$

At $$(0, \pi)$$:

$$\frac{dy}{dx} = \frac{\sin\left(\frac{\pi}{3} + \pi\right)}{\cos \pi - 0 \cdot \cos\left(\frac{\pi}{3} + \pi\right)} = \frac{\sin\left(\frac{4\pi}{3}\right)}{\cos \pi} = \frac{-\frac{\sqrt{3}}{2}}{-1} = \frac{\sqrt{3}}{2}$$

At $$(0, 2\pi)$$:

$$\frac{dy}{dx} = \frac{\sin\left(\frac{\pi}{3} + 2\pi\right)}{\cos 2\pi - 0 \cdot \cos\left(\frac{\pi}{3} + 2\pi\right)} = \frac{\sin\left(\frac{7\pi}{3}\right)}{\cos 2\pi} = \frac{\sin\left(\frac{\pi}{3}\right)}{1} = \frac{\frac{\sqrt{3}}{2}}{1} = \frac{\sqrt{3}}{2}$$

Thus, at every point $$(0, n\pi)$$, the slope of the tangent is $$\frac{\sqrt{3}}{2}$$.

The slope of the normal is the negative reciprocal of the tangent's slope:

$$\text{slope of normal} = -\frac{1}{\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$$

The normal line passes through the point of tangency $$(0, n\pi)$$. However, the given options are linear equations with no constant term, meaning they pass through the origin $$(0, 0)$$. Among the points $$(0, n\pi)$$, only $$(0, 0)$$ lies on the origin. Therefore, we consider the normal at $$(0, 0)$$.

The equation of the normal passing through $$(0, 0)$$ with slope $$-\frac{2}{\sqrt{3}}$$ is:

$$y - 0 = -\frac{2}{\sqrt{3}} (x - 0) \implies y = -\frac{2}{\sqrt{3}} x$$

Multiplying both sides by $$\sqrt{3}$$ to rationalize:

$$\sqrt{3} y = -2x \implies 2x + \sqrt{3} y = 0$$

Comparing with the options:

A. $$2x - \sqrt{3}y = 0$$

B. $$2y - \sqrt{3}x = 0$$

C. $$2y + \sqrt{3}x = 0$$

D. $$2x + \sqrt{3}y = 0$$

The equation $$2x + \sqrt{3} y = 0$$ matches option D.

Verification: At $$(0, 0)$$, the curve passes through the point. The tangent slope at $$(0, 0)$$ is $$\frac{\sqrt{3}}{2}$$, and the normal slope is $$-\frac{2}{\sqrt{3}}$$. Their product is $$\left(\frac{\sqrt{3}}{2}\right) \times \left(-\frac{2}{\sqrt{3}}\right) = -1$$, confirming perpendicularity.

Hence, the correct answer is Option D.