<p>The equation \((\cos p - 1)x^2 + \cos p \cdot x + \sin p = 0\) where \(x\) is a variable, has real roots. Then the interval of possible values of \(p\) is</p>
Step-by-Step Solution
Key Concept: For a quadratic equation with real roots, the discriminant must be non-negative; when the coefficient of x² equals zero, the equation becomes linear and always has real roots. Analyze both cases: quadratic (Δ ≥ 0) and linear.
<p><strong>Step 1:</strong> Identify the cases. If cos p = 1, then the equation becomes x + sin p = 0, which always has a real root x = -sin p. So p = 2nπ is part of the solution.</p><p><strong>Step 2:</strong> If cos p ≠ 1, we have a proper quadratic. For real roots: Δ ≥ 0</p><p>Δ = (cos p)² - 4(cos p - 1)(sin p) ≥ 0</p><p>Δ = cos² p - 4 cos p sin p + 4 sin p ≥ 0</p><p><strong>Step 3:</strong> Rewrite using sin² p + cos² p = 1:</p><p>cos² p - 4 cos p sin p + 4 sin p ≥ 0</p><p>= cos² p + 4 sin p(1 - cos p) ≥ 0</p><p><strong>Step 4:</strong> Note that 1 - cos p ≥ 0 always, and sin p can be positive or negative. Rearranging:</p><p>cos² p + 4 sin p - 4 sin p cos p ≥ 0</p><p>Setting u = cos p, the condition becomes satisfied when sin p ≥ 0 (approximately), which gives p ∈ [2nπ, (2n+1)π]</p><p><strong>Step 5:</strong> Combining both cases: The equation has real roots for p ∈ [2nπ, (2n+1)π], where n ∈ ℤ, or equivalently p ∈ [0, π] (for principal interval).</p><p>∴ Answer: D</p>
Correct Answer: D