If $$e^{\cos^2 x + \cos^4 x + \cos^6 x + \ldots \infty} \log_e 2$$ satisfies the equation $$t^2 - 9t + 8 = 0$$, then the value of $$\frac{2\sin x}{\sin x + \sqrt{3}\cos x}$$, where $$0 < x < \frac{\pi}{2}$$, is equal to

We are given that $$e^{(\cos^2 x + \cos^4 x + \cos^6 x + \ldots \infty) \ln 2}$$ satisfies the equation $$t^2 - 9t + 8 = 0$$.

The infinite geometric series $$\cos^2 x + \cos^4 x + \cos^6 x + \ldots$$ has first term $$\cos^2 x$$ and common ratio $$\cos^2 x$$. Its sum is $$\frac{\cos^2 x}{1 - \cos^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x$$.

So the expression becomes $$e^{\cot^2 x \cdot \ln 2} = 2^{\cot^2 x}$$.

Now, $$t^2 - 9t + 8 = 0$$ gives $$(t - 1)(t - 8) = 0$$, so $$t = 1$$ or $$t = 8$$.

If $$2^{\cot^2 x} = 1$$, then $$\cot^2 x = 0$$, so $$x = \frac{\pi}{2}$$.

If $$2^{\cot^2 x} = 8 = 2^3$$, then $$\cot^2 x = 3$$, so $$\cot x = \sqrt{3}$$ (since $$0 < x < \frac{\pi}{2}$$), giving $$x = \frac{\pi}{6}$$.

Now we evaluate $$\frac{2\sin x}{\sin x + \sqrt{3}\cos x}$$ at $$x = \frac{\pi}{6}$$:

$$\frac{2 \cdot \frac{1}{2}}{\frac{1}{2} + \sqrt{3} \cdot \frac{\sqrt{3}}{2}} = \frac{1}{\frac{1}{2} + \frac{3}{2}} = \frac{1}{2}$$.

At $$x = \frac{\pi}{2}$$: $$\frac{2 \cdot 1}{1 + 0} = 2$$, which is not among the options.

Hence, the correct answer is Option B.