<p>The number of values of \(x\) in the interval \([0, 3\pi]\) satisfying the equation \(2\sin^2 x + 5\sin x - 3 = 0\) is</p>
Step-by-Step Solution
Key Concept: Solve the quadratic equation in sin x by factoring, find the values of sin x, then determine how many solutions exist in [0, 3π] for each valid sine value.
<p><strong>Step 1: Solve the quadratic equation</strong></p><p>We have: 2sin²x + 5sin x - 3 = 0</p><p>Let u = sin x. Then: 2u² + 5u - 3 = 0</p><p>Factoring: (2u - 1)(u + 3) = 0</p><p>This gives: u = 1/2 or u = -3</p><p><strong>Step 2: Determine valid solutions for sin x</strong></p><p>Since sin x must satisfy -1 ≤ sin x ≤ 1:</p><p>• sin x = 1/2 ✓ (valid)</p><p>• sin x = -3 ✗ (invalid, as -3 < -1)</p><p><strong>Step 3: Find all x in [0, 3π] where sin x = 1/2</strong></p><p>In one period [0, 2π], sin x = 1/2 at:</p><p>• x = π/6</p><p>• x = 5π/6</p><p>So there are 2 solutions per period [0, 2π].</p><p><strong>Step 4: Count solutions in [0, 3π]</strong></p><p>The interval [0, 3π] contains 1.5 periods of the sine function.</p><p>In [0, 2π]: x = π/6, 5π/6 (2 solutions)</p><p>In [2π, 3π]: We need sin x = 1/2 in this interval.</p><p>Since sin(x) = sin(x - 2π), solutions in [2π, 3π] occur at:</p><p>• x = 2π + π/6 = 13π/6</p><p>• x = 2π + 5π/6 = 17π/6</p><p>Check: 13π/6 ≈ 2.167π < 3π ✓ and 17π/6 ≈ 2.833π < 3π ✓</p><p><strong>Step 5: Total count</strong></p><p>Solutions: x = π/6, 5π/6, 13π/6, 17π/6</p><p>Total number of solutions = 4</p><p><strong>∴ Answer: D</strong></p>
Correct Answer: D