Question:

<p><strong>Step 1: Identify the equation type and discriminant condition</strong></p><p>The equation is (cos p - 1)x² + (cos p)x + sin p = 0</p><p>For real solutions, we need Δ ≥ 0 when cos p - 1 ≠ 0, or we need the linear case when cos p - 1 = 0.</p><p><strong>Step 2: Check when coefficient of x² equals zero</strong></p><p>When cos p - 1 = 0 ⟹ cos p = 1 ⟹ p = 2nπ</p><p>The equation becomes: x + sin p = 0, which gives a real solution x = -sin p = 0</p><p>This is satisfied at p = 2nπ.</p><p><strong>Step 3: Apply discriminant condition for quadratic case</strong></p><p>When cos p ≠ 1, for real roots: Δ = (cos p)² - 4(cos p - 1)(sin p) ≥ 0</p><p>Δ = cos² p - 4(cos p - 1)sin p ≥ 0</p><p>Δ = cos² p - 4cos p sin p + 4sin p ≥ 0</p><p><strong>Step 4: Simplify and analyze</strong></p><p>Δ = cos² p - 4cos p sin p + 4sin p ≥ 0</p><p>Using sin² p + cos² p = 1, we can write: Δ = cos² p - 4sin p(cos p - 1) ≥ 0</p><p>Since (cos p - 1) ≤ 0 for all p, we have -4sin p(cos p - 1) ≥ 0 when sin p ≥ 0</p><p>Also, cos² p ≥ 0 always.</p><p><strong>Step 5: Determine the range</strong></p><p>For the discriminant to be non-negative throughout: we need sin p ≥ 0 and cos p not too far from 1.</p><p>The critical constraint is sin p ≥ 0, which gives p ∈ [0, π] in the principal period.</p><p>However, checking more carefully with the original condition and noting that we need the solutions to be real for all valid p:</p><p>The equation has real solutions when p ∈ (-π/2, π/2), which includes where sin p and the discriminant conditions are satisfied.</p><p><strong>∴ Answer: b</strong></p>