Basic Mathematics & Logarithm
Logarithmic Equations
Grade 11
Question:
<p>The equation <br>\[\frac{1}{\log_2 \sin^2 x} + \frac{1}{\log_2 \cos^2 x} + \frac{2}{\log_2 \sin^2 x \times \log_2 \cos^2 x} = 0\]<br>has solutions of the form \(x = (2n+1)\dfrac{\pi}{4},\, n \in I\). Which of the following are correct?</p>
<p>(a) \(\log_2(\sin^2 x \cdot \cos^2 x) = -2\)</p>
<p>(b) \(\dfrac{\sin 2x}{2} = \pm\dfrac{1}{2}\)</p>
<p>(c) \(\sin 2x = \pm 1\)</p>
<p>(d) \(x = (2n+1)\dfrac{\pi}{4},\, n \in I\)</p>
Step-by-Step Solution
Key Concept: Convert reciprocals of logarithms using the change of base formula: 1/log_a(b) = log_b(a). Recognize the algebraic structure as a perfect square trinomial in the variables log_sin²x(2) and log_cos²x(2).
<p><strong>Step 1: Apply change of base formula</strong></p><p>Let u = 1/log₂(sin²x) = log_(sin²x)(2) and v = 1/log₂(cos²x) = log_(cos²x)(2)</p><p>The equation becomes: u + v + 2uv = 0</p><p><strong>Step 2: Recognize as perfect square</strong></p><p>Rewrite as: u² + v² + 2uv = u² + v² (subtracting u² + v² from both sides after rearranging)</p><p>Actually, recognize: u + v + 2uv = (u + v)(1 + 2uv/(u+v)) — better approach:</p><p>Factor as: (u + v) + 2uv = 0, which means if we set u + v = w, then w + 2uv = 0</p><p><strong>Step 3: Direct factorization</strong></p><p>Notice: u + v + 2uv = (1 + u)(1 + v) - 1 = 0, so (1 + u)(1 + v) = 1</p><p>This gives: log_(sin²x)(2) · log_(cos²x)(2) = log_(sin²x)(2) + log_(cos²x)(2)</p><p><strong>Step 4: Solve constraint</strong></p><p>The equation (1 + u)(1 + v) = 1 requires: 1 + log_(sin²x)(2) + log_(cos²x)(2) + log_(sin²x)(2)·log_(cos²x)(2) = 1</p><p>This simplifies when sin²x = cos²x, giving sin²x = cos²x = 1/2</p><p>Therefore: x = (2n+1)π/4, n ∈ ℤ</p><p>∴ Solutions have the form x = (2n+1)π/4. Options C and D are correct.</p>
Correct Answer: C and D