1. **State the problem:** Solve the inequality $$\sin^2 x - \sin x \geq 1$$ for real values of $x$. 2. **Rewrite the inequality:** Let $y = \sin x$. The inequality becomes: $$y^2 - y \geq 1$$ 3. **Bring all terms to one side:** $$y^2 - y - 1 \geq 0$$ 4. **Solve the quadratic inequality:** The quadratic expression is $$y^2 - y - 1$$. 5. **Find roots of the quadratic equation:** $$y^2 - y - 1 = 0$$ Using the quadratic formula: $$y = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$$ 6. **Calculate roots:** $$y_1 = \frac{1 - \sqrt{5}}{2} \approx -0.618$$ $$y_2 = \frac{1 + \sqrt{5}}{2} \approx 1.618$$ 7. **Analyze the inequality:** Since the leading coefficient of $y^2$ is positive, the parabola opens upwards. The inequality $$y^2 - y - 1 \geq 0$$ holds when: $$y \leq y_1 \quad \text{or} \quad y \geq y_2$$ 8. **Consider the domain of $\sin x$:** Since $\sin x$ ranges between $-1$ and $1$, the interval $$y \geq y_2 \approx 1.618$$ is impossible. Therefore, the solution reduces to: $$\sin x \leq \frac{1 - \sqrt{5}}{2} \approx -0.618$$ 9. **Find $x$ values satisfying $\sin x \leq -0.618$:** The sine function is periodic with period $2\pi$. The solutions are all $x$ such that: $$\sin x \leq -0.618$$ This corresponds to the arcsin values: $$x \in [\arcsin(-1), \arcsin(-0.618)] \cup [\pi - \arcsin(-0.618), \pi - \arcsin(-1)]$$ Since $\arcsin(-1) = -\frac{\pi}{2}$ and $\arcsin(-0.618) \approx -0.666$ radians, The intervals where $\sin x \leq -0.618$ are: $$x \in \left[-\frac{\pi}{2} + 2k\pi, -0.666 + 2k\pi\right] \cup \left[\pi + 0.666 + 2k\pi, \frac{3\pi}{2} + 2k\pi\right]$$ for all integers $k$. **Final answer:** $$\boxed{\sin x \leq \frac{1 - \sqrt{5}}{2}}$$ which corresponds to the intervals above for $x$.