Under which 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 β ∈ [0, π], cosβ ∈ [–1, 1] and sinβ ≥ 0. If cosβ = 1 (β = 0), equation becomes 0·x² + x + 0 = 0, giving x = 0 (real). If cosβ ≠ 1, discriminant = cos²β – 4(cosβ–1)sinβ = cos²β + 4(1–cosβ)sinβ. Since 1 – cosβ ≥ 0 and sinβ ≥ 0 in [0,π], discriminant ≥ 0 always. So real roots exist for 1 – cosβ ≥ 0, which holds throughout [0, π].