1. **State the problem:** Solve the inequality $$\frac{3\cos(x)}{\sqrt{2} - 2\sin(x)} \geq 0$$ for $x$. 2. **Understand the inequality:** A fraction is non-negative if its numerator and denominator have the same sign (both positive or both negative) or if the numerator is zero. 3. **Analyze numerator:** $3\cos(x) \geq 0$ implies $\cos(x) \geq 0$. 4. **Analyze denominator:** $\sqrt{2} - 2\sin(x) \neq 0$ (denominator cannot be zero). 5. **Find when denominator is zero:** $$\sqrt{2} - 2\sin(x) = 0 \implies 2\sin(x) = \sqrt{2} \implies \sin(x) = \frac{\sqrt{2}}{2}$$ This happens at $x = \frac{\pi}{4} + 2k\pi$ and $x = \frac{3\pi}{4} + 2k\pi$, $k \in \mathbb{Z}$. 6. **Determine sign of denominator:** - For $\sin(x) < \frac{\sqrt{2}}{2}$, denominator $>0$. - For $\sin(x) > \frac{\sqrt{2}}{2}$, denominator $<0$. 7. **Determine sign of numerator:** $\cos(x) \geq 0$ means $x \in [-\frac{\pi}{2} + 2k\pi, \frac{\pi}{2} + 2k\pi]$. 8. **Combine conditions for fraction $\geq 0$:** - Case 1: numerator $\geq 0$ and denominator $>0$. - Case 2: numerator $\leq 0$ and denominator $<0$. 9. **Case 1:** $$\cos(x) \geq 0 \quad \text{and} \quad \sin(x) < \frac{\sqrt{2}}{2}$$ 10. **Case 2:** $$\cos(x) \leq 0 \quad \text{and} \quad \sin(x) > \frac{\sqrt{2}}{2}$$ 11. **Include points where numerator is zero:** $$\cos(x) = 0 \implies x = \frac{\pi}{2} + k\pi$$ At these points, fraction equals zero (allowed). 12. **Final solution:** $$\boxed{\left\{ x \mid (\cos(x) \geq 0 \text{ and } \sin(x) < \frac{\sqrt{2}}{2}) \text{ or } (\cos(x) \leq 0 \text{ and } \sin(x) > \frac{\sqrt{2}}{2}) \right\} \cup \left\{ x \mid \cos(x) = 0 \right\}}$$ This describes all $x$ where the original inequality holds, excluding points where denominator is zero.