Trigonometry & Inverse Trigonometry
General Solutions of Trigonometric Equations
Grade 11
Question:
<p>The general solution of \(\sin x + \cos x = 1\) is given by</p>
<p>(a) \(2n\pi\)</p>
<p>(b) \(2n\pi + \dfrac{\pi}{2}\)</p>
<p>(c) \(n\pi + (-1)^n \dfrac{\pi}{4} - \dfrac{\pi}{4}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Rewrite sin x + cos x using the sum-to-product form √2·sin(x + π/4) = 1, then solve sin(x + π/4) = 1/√2 to get two families of solutions that must be reconciled.
<p><strong>Step 1:</strong> Convert to standard form using auxiliary angle method:</p><p>sin x + cos x = √2·sin(x + π/4)</p><p>So the equation becomes: √2·sin(x + π/4) = 1</p><p>⟹ sin(x + π/4) = 1/√2 = sin(π/4)</p><p></p><p><strong>Step 2:</strong> General solution of sin θ = sin α is θ = nπ + (-1)ⁿα</p><p>x + π/4 = nπ + (-1)ⁿ(π/4)</p><p></p><p><strong>Step 3:</strong> This gives two cases:</p><p>• When n is even: x + π/4 = 2mπ + π/4 ⟹ x = 2mπ</p><p>• When n is odd: x + π/4 = (2m+1)π - π/4 ⟹ x = π/2 + 2mπ</p><p></p><p><strong>Step 4:</strong> Verify by substitution:</p><p>• x = 2mπ: sin(2mπ) + cos(2mπ) = 0 + 1 = 1 ✓</p><p>• x = π/2 + 2mπ: sin(π/2) + cos(π/2) = 1 + 0 = 1 ✓</p><p></p><p>Both families work! ∴ General solution: <strong>x = 2nπ or x = π/2 + 2nπ, n ∈ ℤ</strong></p><p>This can be written as: x = 2nπ, 2nπ + π/2 where n ∈ ℤ</p>
Correct Answer: C