Let's solve the given trigonometric equation step by step:
Original equation:
$$2 + 2\cos(\pi - 2x) + \sqrt{8}\sin x = \sqrt{6} + \sqrt{12}\sin x$$
Using the property $$\cos(\pi - \alpha) = -\cos(\alpha)$$:
$$2 - 2\cos(2x) + \sqrt{8}\sin x = \sqrt{6} + \sqrt{12}\sin x$$
Rewrite the radicals:
$$2 - 2\cos(2x) + 2\sqrt{2}\sin x = \sqrt{6} + 2\sqrt{3}\sin x$$
Using the double angle formula $$\cos(2x) = 1 - 2\sin^2(x)$$:
$$2 - 2(1 - 2\sin^2(x)) + 2\sqrt{2}\sin x = \sqrt{6} + 2\sqrt{3}\sin x$$
Simplify:
$$2 - 2 + 4\sin^2(x) + 2\sqrt{2}\sin x = \sqrt{6} + 2\sqrt{3}\sin x$$
$$4\sin^2(x) + 2\sqrt{2}\sin x - 2\sqrt{3}\sin x = \sqrt{6}$$
$$4\sin^2(x) + (2\sqrt{2} - 2\sqrt{3})\sin x - \sqrt{6} = 0$$
Let $$y = \sin x$$:
$$4y^2 + 2(\sqrt{2} - \sqrt{3})y - \sqrt{6} = 0$$
Using the quadratic formula:
$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
$$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm \sqrt{(2(\sqrt{2} - \sqrt{3}))^2 - 4(4)(-\sqrt{6})}}{2(4)}$$
$$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm \sqrt{4(2 - 2\sqrt{6} + 3) + 16\sqrt{6}}}{8}$$
$$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm \sqrt{4(5 - 2\sqrt{6}) + 16\sqrt{6}}}{8}$$
$$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm \sqrt{20 - 8\sqrt{6} + 16\sqrt{6}}}{8}$$
$$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm \sqrt{20 + 8\sqrt{6}}}{8}$$
$$y = \frac{-2(\sqrt{2} - \sqrt{3}) \pm 2\sqrt{5 + 2\sqrt{6}}}{8}$$
$$y = \frac{-(\sqrt{2} - \sqrt{3}) \pm \sqrt{(\sqrt{2} + \sqrt{3})^2}}{4}$$
$$y = \frac{-\sqrt{2} + \sqrt{3} \pm (\sqrt{2} + \sqrt{3})}{4}$$
Case 1: $$y = \frac{-\sqrt{2} + \sqrt{3} + \sqrt{2} + \sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$$
Case 2: $$y = \frac{-\sqrt{2} + \sqrt{3} - \sqrt{2} - \sqrt{3}}{4} = \frac{-2\sqrt{2}}{4} = -\frac{\sqrt{2}}{2}$$
Thus, $$\sin x = \frac{\sqrt{3}}{2}$$ or $$\sin x = -\frac{\sqrt{2}}{2}$$.
For $$\sin x = \frac{\sqrt{3}}{2}$$, the solutions are:
$$x = \frac{\pi}{3} + 2\pi k, \quad x = \frac{2\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}$$
For $$\sin x = -\frac{\sqrt{2}}{2}$$, the solutions are:
$$x = -\frac{\pi}{4} + 2\pi k = \frac{7\pi}{4} + 2\pi k, \quad x = \frac{5\pi}{4} + 2\pi k, \quad k \in \mathbb{Z}$$
Final Answer: The final answer is:
$$x = \frac{\pi}{3} + 2\pi k, \quad x = \frac{2\pi}{3} + 2\pi k, \quad x = \frac{5\pi}{4} + 2\pi k, \quad x = \frac{7\pi}{4} + 2\pi k, \quad k \in \mathbb{Z}$$