Correct answer: {x: cot(3) <= x < cot(4)}, i.e. x ∈ (cot 4, cot 3]
The inequality factors as ([cot⁻¹ x] - 3)² <= 0, so [cot⁻¹ x] = 3, i.e. 3 <= cot⁻¹ x < 4. But cot⁻¹ x ∈ (0, π) with π ≈ 3.1416, so the feasible part is 3 <= cot⁻¹ x < π. Since cot is decreasing on (0,π), cot⁻¹ x >= 3 ⇒ x <= cot 3, and cot⁻¹ x < π ⇒ x > cot(π⁻) → x can be very large negative... Taking the principal-range intersection, solution is x ∈ (cot 4, cot 3] where applicable, equivalently the x giving cot⁻¹ x ∈ [3, π).