<p>The number of solutions of the equation \(2\sin^3\alpha - 7\sin^2\alpha + 7\sin\alpha - 2 = 0\) in \([0, 2\pi]\) is</p>
Step-by-Step Solution
Key Concept: Factor the cubic as (sin α - 1)(2sin²α - 5sin α + 2) = 0, then solve each factor separately by substituting sin α = t ∈ [-1, 1], and count valid solutions in [0, 2π].
<p><strong>Step 1:</strong> Substitute t = sin α where t ∈ [-1, 1]. The equation becomes: 2t³ - 7t² + 7t - 2 = 0</p><p><strong>Step 2:</strong> Check t = 1: 2(1) - 7(1) + 7(1) - 2 = 0 ✓. So (t - 1) is a factor.</p><p><strong>Step 3:</strong> Divide: 2t³ - 7t² + 7t - 2 = (t - 1)(2t² - 5t + 2)</p><p><strong>Step 4:</strong> Factor the quadratic: 2t² - 5t + 2 = (2t - 1)(t - 2) = 0, giving t = 1/2 or t = 2</p><p><strong>Step 5:</strong> Solutions for sin α:</p><ul><li><strong>sin α = 1:</strong> α = π/2 (1 solution in [0, 2π])</li><li><strong>sin α = 1/2:</strong> α = π/6, 5π/6 (2 solutions in [0, 2π])</li><li><strong>sin α = 2:</strong> No solution (2 > 1, outside range of sine)</li></ul><p><strong>Step 6:</strong> Total number of solutions = 1 + 2 = <strong>3</strong></p><p>∴ Answer: C</p>
Correct Answer: C