Under which one of the following conditions does the equation (cosβ – 1)x² + (cosβ)x + sinβ = 0 in x have a real root for β ∈ [0, π]?

For the quadratic (cosβ – 1)x² + (cosβ)x + sinβ = 0 to have real roots, the discriminant must satisfy cos²β – 4(cosβ – 1)sinβ ≥ 0. For β ∈ [0, π], sinβ ≥ 0 and (cosβ – 1) ≤ 0, so –4(cosβ – 1)sinβ = 4(1 – cosβ)sinβ ≥ 0. Adding cos²β ≥ 0 keeps the discriminant non-negative throughout. The equation becomes linear (still having a real root) when cosβ = 1, i.e., 1 – cosβ = 0. Thus the necessary condition covering both quadratic and degenerate-linear cases is 1 – cosβ ≥ 0, which holds for all β ∈ [0, π].