Trigonometry & Inverse Trigonometry
General Solution of Trigonometric Equations
Grade 11
Question:
<p>The general solution of <strong>sin 2θ sec θ + √3 tan θ = 0</strong>.</p>
<p>(a) <strong>θ = nπ + (-1)ⁿ⁺¹ π/3</strong>, <strong>θ = nπ</strong>; <strong>n ∈ ℤ</strong></p>
<p>(b) <strong>θ = nπ</strong>, <strong>n ∈ ℤ</strong></p>
<p>(c) <strong>θ = nπ/2</strong>, <strong>n ∈ ℤ</strong></p>
<p>(d) <strong>θ = nπ + (-1)ⁿ⁺¹ π/4</strong>, <strong>n ∈ ℤ</strong></p>
Step-by-Step Solution
Key Concept: Factor the equation and recognize that one factor yields valid solutions while the other is impossible due to the range of sine.
<p><strong>Step 1:</strong> Rewrite: <strong>sin 2θ sec θ + √3 tan θ = 0</strong> as <strong>sin θ(sin θ + √3) sec θ = 0</strong> (using <strong>sin 2θ = 2 sin θ cos θ</strong>).</p><p><strong>Step 2:</strong> Either <strong>sin θ = 0</strong> or <strong>sin θ + √3 = 0</strong>.</p><p><strong>Step 3:</strong> From <strong>sin θ = 0</strong>: <strong>θ = nπ</strong>, <strong>n ∈ ℤ</strong>.</p><p><strong>Step 4:</strong> From <strong>sin θ = -√3</strong>: This is impossible since <strong>|sin θ| ≤ 1 < √3</strong>.</p><p><strong>Step 5:</strong> Also, <strong>sec θ ≠ 0</strong> is automatically satisfied where sin θ = 0 (since cos θ = ±1).</p><p><strong>∴ Answer is (b): θ = nπ, n ∈ ℤ</strong></p>
Correct Answer: B