1. **State the problem:** Find the interval of $x$ such that $$\left|\,|x-3|-2\right| \geq 1.$$\n\n2. **Understand the expression:** The expression involves nested absolute values. Let $y = |x-3|$. Then the inequality becomes $$|y - 2| \geq 1.$$\n\n3. **Solve the inner inequality:** For $|y - 2| \geq 1$, this means $$y - 2 \geq 1 \quad \text{or} \quad y - 2 \leq -1,$$ which simplifies to $$y \geq 3 \quad \text{or} \quad y \leq 1.$$\n\n4. **Recall $y = |x-3|$:** So we have two cases:\n- Case 1: $$|x-3| \geq 3,$$\n- Case 2: $$|x-3| \leq 1.$$\n\n5. **Solve Case 1:** $$|x-3| \geq 3$$ means $$x-3 \geq 3 \quad \text{or} \quad x-3 \leq -3,$$ which gives $$x \geq 6 \quad \text{or} \quad x \leq 0.$$\n\n6. **Solve Case 2:** $$|x-3| \leq 1$$ means $$-1 \leq x-3 \leq 1,$$ which gives $$2 \leq x \leq 4.$$\n\n7. **Combine the solution sets:** The original inequality holds when $$x \leq 0, \quad 2 \leq x \leq 4, \quad \text{or} \quad x \geq 6.$$\n\n8. **Interpretation:** The solution intervals are $$(-\infty, 0] \cup [2,4] \cup [6, \infty).$$\n\nThis matches the graph where the function $y = |\,|x-3|-2|$ is greater than or equal to 1 at these intervals.\n\n**Final answer:** $$x \in (-\infty, 0] \cup [2,4] \cup [6, \infty).$$