Trigonometry & Inverse Trigonometry
Trigonometric equations
Grade 11
Question:
<p>If \(2\sin^2\theta + 2\sqrt{2} = 3\csc^2\theta\), where \(\theta \in (0, \pi)\), then:</p>
<p>Number of real solutions is 2.</p>
<p>Number of real solution is 4.</p>
<p>Sum of all solutions is \(\pi\).</p>
<p>Sum of all solutions is \(4\pi\).</p>
Step-by-Step Solution
Key Concept: Convert the equation to a single trigonometric variable by substituting sin²θ = x, then solve the resulting quadratic. The constraint θ ∈ (0,π) ensures sinθ > 0, which determines valid solutions.
<p><strong>Step 1:</strong> Rewrite using csc²θ = 1/sin²θ:</p><p>2sin²θ + 2√2 = 3/sin²θ</p><p><strong>Step 2:</strong> Let x = sin²θ where 0 < x ≤ 1 (since θ ∈ (0,π), sinθ > 0):</p><p>2x + 2√2 = 3/x</p><p><strong>Step 3:</strong> Multiply by x:</p><p>2x² + 2√2·x = 3</p><p>2x² + 2√2·x - 3 = 0</p><p><strong>Step 4:</strong> Using the quadratic formula:</p><p>x = (-2√2 ± √(8 + 24))/4 = (-2√2 ± √32)/4 = (-2√2 ± 4√2)/4</p><p><strong>Step 5:</strong> x = (2√2)/4 = √2/2 or x = (-6√2)/4 (rejected, negative)</p><p><strong>Step 6:</strong> So sin²θ = 1/2, giving sinθ = 1/√2 (positive in (0,π))</p><p><strong>Step 7:</strong> Therefore θ = π/4 or θ = 3π/4 (both in (0,π))</p><p>∴ Answer: AC</p>
Correct Answer: AC