The solutions of the equation $$\begin{vmatrix} 1 + \sin^2 x & \sin^2 x & \sin^2 x \\ \cos^2 x & 1 + \cos^2 x & \cos^2 x \\ 4\sin 2x & 4\sin 2x & 1 + 4\sin 2x \end{vmatrix} = 0$$, $$(0 < x < \pi)$$, are:

We apply column operations to simplify the determinant. Let $$s = \sin^2 x$$, $$c = \cos^2 x$$, and $$d = 4\sin 2x$$. The determinant is $$\begin{vmatrix} 1+s & s & s \\ c & 1+c & c \\ d & d & 1+d \end{vmatrix}$$.

Apply $$C_1 \to C_1 - C_2$$ and $$C_3 \to C_3 - C_2$$: $$\begin{vmatrix} 1 & s & 0 \\ -1 & 1+c & -1 \\ 0 & d & 1 \end{vmatrix}$$.

Expanding along the first row: $$1 \cdot \begin{vmatrix} 1+c & -1 \\ d & 1 \end{vmatrix} - s \cdot \begin{vmatrix} -1 & -1 \\ 0 & 1 \end{vmatrix} + 0 = (1+c+d) - s(-1-0) = 1 + c + d + s$$.

Since $$s + c = \sin^2 x + \cos^2 x = 1$$, this simplifies to $$1 + 1 + d = 2 + 4\sin 2x$$.

Setting this to zero: $$2 + 4\sin 2x = 0$$, so $$\sin 2x = -\frac{1}{2}$$.

For $$0 < x < \pi$$, we have $$0 < 2x < 2\pi$$. The solutions of $$\sin 2x = -\frac{1}{2}$$ in this range are $$2x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$$ and $$2x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$$.

Therefore $$x = \frac{7\pi}{12}$$ and $$x = \frac{11\pi}{12}$$.

The answer is Option D: $$\frac{7\pi}{12}, \frac{11\pi}{12}$$.