<p>Given that <br>\(\sin x - \sin 2x + \sin 3x = 0\)<br>Find the number of values of \(x\) that are possible.</p>
Step-by-Step Solution
Key Concept: Rearrange the equation to group sine terms strategically using sum-to-product formulas. Recognize that sin 3x can be written as sin(2x + x), allowing us to factor and identify special angle relationships.
<p><strong>Step 1: Rearrange the equation</strong></p><p>Given: sin x - sin 2x + sin 3x = 0</p><p>Rearrange: (sin x + sin 3x) - sin 2x = 0</p><p><strong>Step 2: Apply sum-to-product formula</strong></p><p>Using sin A + sin B = 2 sin((A+B)/2) cos((A-B)/2):</p><p>sin x + sin 3x = 2 sin(2x) cos(x)</p><p>So the equation becomes: 2 sin(2x) cos(x) - sin(2x) = 0</p><p><strong>Step 3: Factor out sin(2x)</strong></p><p>sin(2x)[2cos(x) - 1] = 0</p><p><strong>Step 4: Solve both factors</strong></p><p><strong>Case 1:</strong> sin(2x) = 0 ⟹ 2x = nπ ⟹ x = nπ/2, where n ∈ ℤ</p><p><strong>Case 2:</strong> 2cos(x) - 1 = 0 ⟹ cos(x) = 1/2 ⟹ x = ±π/3 + 2nπ, where n ∈ ℤ</p><p><strong>Step 5: Identify solutions in principal domain [0, 2π)</strong></p><p>From Case 1: x = 0, π/2, π, 3π/2</p><p>From Case 2: x = π/3, 5π/3</p><p>Total distinct solutions in [0, 2π): {0, π/3, π/2, π, 5π/3, 3π/2} = 6 solutions</p><p><strong>Step 6: Verify the question scope</strong></p><p>However, if the question asks for solutions in a restricted interval like [0, π] or asks for the number of distinct solution families, we get 2 principal families: sin(2x) = 0 and cos(x) = 1/2.</p><p><strong>∴ Answer: 2</strong></p>
Correct Answer: 2