<p>Given \(\sin 3x = \cos 2x\), the number of solutions in \(x \in \left(\dfrac{\pi}{2}, \pi\right)\) is:</p>
Step-by-Step Solution
Key Concept: Convert sin 3x = cos 2x to sin 3x = sin(π/2 - 2x), then use the general solution for sin A = sin B: A = nπ + (-1)^n B to find all values in the given interval.
<p><strong>Step 1:</strong> Convert to same trigonometric function: sin 3x = cos 2x = sin(π/2 - 2x)</p><p><strong>Step 2:</strong> Apply general solution sin A = sin B ⟹ A = nπ + (-1)^n B</p><p>Case 1: 3x = nπ + (-1)^n(π/2 - 2x)</p><p><strong>Step 3 (Subcase n even):</strong> 3x = nπ + π/2 - 2x ⟹ 5x = nπ + π/2 ⟹ x = (2n+1)π/10</p><p>For x ∈ (π/2, π): π/2 < (2n+1)π/10 < π ⟹ 5 < 2n+1 < 10 ⟹ n = 2, 3, 4</p><p>Solutions: x = 5π/10 = π/2 (boundary, excluded), 7π/10, 9π/10 ✓</p><p><strong>Step 4 (Subcase n odd):</strong> 3x = nπ - π/2 + 2x ⟹ x = nπ - π/2</p><p>For x ∈ (π/2, π): π/2 < nπ - π/2 < π ⟹ 1 < n < 3/2 ⟹ n = 1</p><p>Solution: x = π/2 (boundary, excluded)</p><p><strong>Step 5:</strong> Valid solutions in (π/2, π): x = 7π/10, 9π/10</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: A