1. State the problem: Solve the trigonometric equation $$\sin x = \frac{1}{2}$$ for $$x$$ in the interval $$[0, 2\pi]$$. 2. Recall that $$\sin x = \frac{1}{2}$$ at angles where $$x = \frac{\pi}{6}$$ and $$x = \frac{5\pi}{6}$$ within one complete cycle $$[0, 2\pi]$$. 3. Therefore, the solutions to $$\sin x = \frac{1}{2}$$ on the interval $$[0, 2\pi]$$ are $$x = \frac{\pi}{6}$$ and $$x = \frac{5\pi}{6}$$. 4. Final answer: $$x = \frac{\pi}{6}, \frac{5\pi}{6}$$