All possible values of $$\theta \in [0, 2\pi]$$ for which $$\sin 2\theta + \tan 2\theta > 0$$ lie in:

We need to find all $$\theta \in [0, 2\pi]$$ for which $$\sin 2\theta + \tan 2\theta > 0$$.

First, note that $$\tan 2\theta$$ is undefined when $$\cos 2\theta = 0$$, i.e., at $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$. We exclude these values.

For all other $$\theta$$, we combine the terms: $$\sin 2\theta + \tan 2\theta = \sin 2\theta + \frac{\sin 2\theta}{\cos 2\theta} = \sin 2\theta \cdot \frac{\cos 2\theta + 1}{\cos 2\theta}$$.

Using the double angle identity $$\cos 2\theta + 1 = 2\cos^2\theta$$, the expression becomes $$\frac{2\sin 2\theta \cos^2\theta}{\cos 2\theta}$$.

Now $$\cos^2\theta = 0$$ only at $$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$$. At these points, $$\sin 2\theta = \sin\pi = 0$$ and $$\sin 3\pi = 0$$ respectively, and $$\tan 2\theta = \tan\pi = 0$$ and $$\tan 3\pi = 0$$ respectively. So $$\sin 2\theta + \tan 2\theta = 0$$ at these points, and the inequality is not satisfied. We exclude them.

For all remaining $$\theta$$, we have $$\cos^2\theta > 0$$, so $$2\cos^2\theta > 0$$. Therefore the sign of the expression depends only on $$\frac{\sin 2\theta}{\cos 2\theta} = \tan 2\theta$$.

The inequality reduces to $$\tan 2\theta > 0$$. The tangent function is positive in the first and third quadrants: $$\tan 2\theta > 0$$ when $$2\theta \in \left(0, \frac{\pi}{2}\right) \cup \left(\pi, \frac{3\pi}{2}\right) \cup \left(2\pi, \frac{5\pi}{2}\right) \cup \left(3\pi, \frac{7\pi}{2}\right)$$.

Dividing all bounds by 2: $$\theta \in \left(0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right) \cup \left(\frac{3\pi}{2}, \frac{7\pi}{4}\right)$$.

Therefore, the solution set is $$\left(0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right) \cup \left(\frac{3\pi}{2}, \frac{7\pi}{4}\right)$$.