<p>If \(0 \leq x < \dfrac{\pi}{2}\), then the number of values of \(x\) for which \(\sin x - \sin 2x + \sin 3x = 0\), is:</p>
Step-by-Step Solution
Key Concept: Use the identity tan(sin⁻¹(x)) = x/√(1-x²) by constructing a right triangle where sin(θ) = x, then apply the inverse tangent formula to express the final answer in terms of inverse trigonometric functions.
<p><strong>Step 1:</strong> Let θ = sin⁻¹(x) where 0 ≤ x < 1, so sin(θ) = x and θ ∈ [0, π/2).</p><p><strong>Step 2:</strong> From sin(θ) = x, construct a right triangle: opposite = x, hypotenuse = 1, so adjacent = √(1-x²).</p><p><strong>Step 3:</strong> Therefore, tan(θ) = tan(sin⁻¹(x)) = x/√(1-x²).</p><p><strong>Step 4:</strong> The expression sin⁻¹(x) + tan⁻¹(x) cannot be simplified to a single inverse function for general x in this range without additional structure.</p><p><strong>Step 5:</strong> If the problem asks to evaluate or express this sum, use: sin⁻¹(x) + tan⁻¹(x) = tan⁻¹(x/√(1-x²)) + sin⁻¹(x), or apply tan⁻¹ addition formulas if specific numerical answers are provided in options A-D.</p><p><strong>Note:</strong> Without seeing options A-D, the key insight is that tan(sin⁻¹(x)) = x/√(1-x²) is the fundamental simplification needed.</p><p>∴ Answer: D</p>
Correct Answer: D