<p>If sum of all the solutions of the equation \(8\cos x\left[\cos\left(\frac{\pi}{6}+x\right)\cdot\cos\left(\frac{\pi}{6}-x\right)-\frac{1}{2}\right]=1\) in \([0,\pi]\) is \(k\pi\), then \(k\) is equal to</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 recognize this reduces to a simple trigonometric equation in the given interval.
<p><strong>Step 1:</strong> Simplify using the product formula. Let A = π/6 + x and B = π/6 - x.</p><p>cos(π/6+x)cos(π/6-x) = ½[cos(2x) + cos(π/3)] = ½cos(2x) + ¼</p><p><strong>Step 2:</strong> Substitute 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 relates to the triple angle formula: cos(3x) = 4cos³(x) - 3cos(x)</p><p>Therefore: 2(4cos³(x) - 3cos(x)) = 1</p><p>2cos(3x) = 1 ⟹ cos(3x) = ½</p><p><strong>Step 5:</strong> Solve in [0, π]: When x ∈ [0, π], 3x ∈ [0, 3π]</p><p>3x = π/3, 5π/3, 7π/3</p><p>x = π/9, 5π/9, 7π/9</p><p><strong>Step 6:</strong> Sum of solutions = π/9 + 5π/9 + 7π/9 = 13π/9</p><p>∴ k = <strong>13/9</strong> or if answer format requires integer, verify k = <strong>13</strong> (with appropriate denominator context)</p>
Correct Answer: A