Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>Given the equation \(8\cos x\left[\cos\left(\dfrac{\pi}{6}+x\right)\cdot\cos\left(\dfrac{\pi}{6}-x\right)-\dfrac{1}{2}\right]=1\), find the sum of all solutions in \([0, \pi]\) and express in the form \(k\pi\). What is \(k\)?</p>
<p>\(k = \dfrac{13}{9}\)</p>
<p>\(k = \dfrac{9}{13}\)</p>
<p>\(k = \dfrac{5}{9}\)</p>
<p>\(k = \dfrac{7}{9}\)</p>
Step-by-Step Solution
Key Concept: Use the product-to-sum formula cos(A)cos(B) = ½[cos(A-B) + cos(A+B)] to simplify cos(π/6+x)cos(π/6-x), then reduce to a standard trigonometric equation.
<p><strong>Step 1:</strong> Apply product formula to cos(π/6+x)cos(π/6-x):</p><p>cos(π/6+x)cos(π/6-x) = ½[cos(2x) + cos(π/3)] = ½[cos(2x) + ½] = ½cos(2x) + ¼</p><p><strong>Step 2:</strong> Substitute back into the original equation:</p><p>8cos(x)[½cos(2x) + ¼ - ½] = 1</p><p>8cos(x)[½cos(2x) - ¼] = 1</p><p>4cos(x)cos(2x) - 2cos(x) = 1</p><p><strong>Step 3:</strong> Use cos(2x) = 2cos²(x) - 1:</p><p>4cos(x)(2cos²(x) - 1) - 2cos(x) = 1</p><p>8cos³(x) - 4cos(x) - 2cos(x) = 1</p><p>8cos³(x) - 6cos(x) - 1 = 0</p><p><strong>Step 4:</strong> Recognize this as 2(4cos³(x) - 3cos(x)) = 1, and recall cos(3x) = 4cos³(x) - 3cos(x):</p><p>2cos(3x) = 1 → cos(3x) = ½</p><p><strong>Step 5:</strong> Solve cos(3x) = ½ in [0, π]:</p><p>3x ∈ [0, 3π], so 3x = π/3 or 3x = 5π/3</p><p>x = π/9 or x = 5π/9</p><p><strong>Step 6:</strong> Sum of solutions: π/9 + 5π/9 = 6π/9 = 2π/3</p><p>∴ k = <strong>2/3</strong> (or if answer format requires k where sum = kπ, then k = 2/3)</p>
Correct Answer: A