<p>The number of solutions of \(\sin 3x = \cos 2x\), in the interval \(\left(\dfrac{\pi}{2}, \pi\right)\) 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 of sin A = sin B, which gives A = nπ + (-1)ⁿB. Apply the constraint (π/2, π) to count valid solutions.
<p><strong>Step 1:</strong> Rewrite the equation using cos 2x = sin(π/2 - 2x):<br/>sin 3x = sin(π/2 - 2x)</p><p><strong>Step 2:</strong> Apply the general solution sin A = sin B:<br/>3x = nπ + (-1)ⁿ(π/2 - 2x), where n ∈ ℤ</p><p><strong>Step 3:</strong> <strong>Case 1 (n even, n = 2k):</strong><br/>3x = 2kπ + π/2 - 2x<br/>5x = 2kπ + π/2<br/>x = (4kπ + π)/10 = π(4k + 1)/10</p><p><strong>For x ∈ (π/2, π):</strong><br/>π/2 < π(4k + 1)/10 < π<br/>5 < 4k + 1 < 10<br/>4 < 4k < 9<br/>1 < k < 2.25<br/>So k = 2, giving x = 9π/10 ✓</p><p><strong>Step 4:</strong> <strong>Case 2 (n odd, n = 2k + 1):</strong><br/>3x = (2k + 1)π - (π/2 - 2x)<br/>3x = (2k + 1)π - π/2 + 2x<br/>x = (2k + 1)π - π/2 = π(4k + 1)/2</p><p><strong>For x ∈ (π/2, π):</strong><br/>π/2 < π(4k + 1)/2 < π<br/>1 < 4k + 1 < 2<br/>0 < 4k < 1<br/>0 < k < 0.25<br/>No integer solutions</p><p><strong>Step 5:</strong> Verify x = 9π/10:<br/>sin(27π/10) = sin(27π/10 - 2π) = sin(7π/10)<br/>cos(18π/10) = cos(9π/5) = cos(9π/5 - 2π) = cos(-π/5) = cos(π/5)<br/>Check: sin(7π/10) = sin(π - 3π/10) = sin(3π/10) and cos(π/5) = sin(π/2 - π/5) = sin(3π/10) ✓</p><p>∴ <strong>The number of solutions is 1</strong></p>
Correct Answer: A