Trigonometry & Inverse Trigonometry
General Solutions of Trigonometric Equations
Grade 11
Question:
<p>The equation \(\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x\) has solution</p>
<p>(a) \(n\pi\)</p>
<p>(b) \(\dfrac{n\pi}{2} + \dfrac{\pi}{8}\)</p>
<p>(c) (none given clearly)</p>
<p>(d) (none given clearly)</p>
Step-by-Step Solution
Key Concept: Rearrange the equation by grouping sine and cosine terms separately, then use sum-to-product formulas to factor. The equation reduces to a simple trigonometric identity when you recognize that (sin x + sin 3x) and (cos x + cos 3x) share a common factor.
<p><strong>Step 1:</strong> Rearrange the equation as:</p><p>(sin x + sin 3x) - 3sin 2x = (cos x + cos 3x) - 3cos 2x</p><p><strong>Step 2:</strong> Apply sum-to-product formulas:</p><p>sin x + sin 3x = 2sin 2x cos x</p><p>cos x + cos 3x = 2cos 2x cos x</p><p><strong>Step 3:</strong> Substitute:</p><p>2sin 2x cos x - 3sin 2x = 2cos 2x cos x - 3cos 2x</p><p>sin 2x(2cos x - 3) = cos 2x(2cos x - 3)</p><p><strong>Step 4:</strong> Case 1: If 2cos x - 3 ≠ 0, then sin 2x = cos 2x</p><p>tan 2x = 1 ⟹ 2x = π/4 + nπ ⟹ x = π/8 + nπ/2, n ∈ ℤ</p><p><strong>Step 5:</strong> Case 2: If 2cos x - 3 = 0 ⟹ cos x = 3/2 (impossible since |cos x| ≤ 1)</p><p>∴ Answer: <strong>x = π/8 + nπ/2, n ∈ ℤ</strong> or equivalently <strong>x = (4n+1)π/8, n ∈ ℤ</strong></p>
Correct Answer: B