Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>If <span class="math">\frac{\sin^2 2x + 4\sin 4x - 4\sin 2x \times \cos 2x}{4 - \sin^2 2x - 4\sin 2x} = \frac{1}{9}</span> and <span class="math">0 < x < \pi</span>, then the value of <span class="math">x</span> is:</p>
<p>(a) <span class="math">\frac{\pi}{3}</span></p>
<p>(b) <span class="math">\frac{\pi}{6}</span></p>
<p>(c) <span class="math">\frac{2\pi}{3}</span></p>
<p>(d) <span class="math">\frac{5\pi}{6}</span></p>
Step-by-Step Solution
Key Concept: Simplify the complex trigonometric expression using double angle formulas and solve the resulting equation systematically.
<p><strong>Solution:</strong> Simplify the given equation by substituting <span class="math">\sin 2x</span> and <span class="math">\cos 2x</span> terms. After algebraic manipulation and solving the resulting trigonometric equation, the values satisfying the constraint <span class="math">0 < x < \pi</span> are found.</p><p>∴ Answer is (b) and (d).</p>
Correct Answer: B, D