1. **State the problem:** Solve the inequality $$\sin^2 x - \sin x \geq 0$$. 2. **Rewrite the inequality:** Factor the left side as a quadratic in terms of $\sin x$: $$\sin^2 x - \sin x = \sin x (\sin x - 1) \geq 0$$. 3. **Analyze the product:** The product $\sin x (\sin x - 1)$ is greater than or equal to zero when both factors are non-negative or both are non-positive. 4. **Case 1: Both factors non-negative:** $$\sin x \geq 0 \quad \text{and} \quad \sin x - 1 \geq 0 \implies \sin x \geq 1$$ Since $\sin x \leq 1$ always, this means: $$\sin x = 1$$ which occurs at $$x = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$$. 5. **Case 2: Both factors non-positive:** $$\sin x \leq 0 \quad \text{and} \quad \sin x - 1 \leq 0$$ The second inequality is always true since $\sin x - 1 \leq 0$ for all $x$. So this reduces to: $$\sin x \leq 0$$ which means $x$ is in intervals where sine is negative or zero: $$x \in [\pi + 2k\pi, 2\pi + 2k\pi], \quad k \in \mathbb{Z}$$. 6. **Combine solutions:** The solution set is: $$\{x \mid \sin x = 1\} \cup \{x \mid \sin x \leq 0\}$$ which is: $$x = \frac{\pi}{2} + 2k\pi \quad \text{or} \quad x \in [\pi + 2k\pi, 2\pi + 2k\pi], \quad k \in \mathbb{Z}$$. **Final answer:** $$\boxed{x = \frac{\pi}{2} + 2k\pi \quad \text{or} \quad x \in [\pi + 2k\pi, 2\pi + 2k\pi], \quad k \in \mathbb{Z}}$$