<p>The sum of solutions in \((0, 2\pi)\) of the equation \(\cos x \cos\!\left(\dfrac{\pi}{3} - x\right)\cos\!\left(\dfrac{\pi}{3} + x\right) = \dfrac{1}{4}\) is:</p>
Step-by-Step Solution
Key Concept: Use the product-to-sum identity on the last two cosines: cos(A-B)cos(A+B) = cos²A - sin²B, then apply double angle formulas to simplify to a single cosine equation.
<p><strong>Step 1:</strong> Apply product formula to cos(π/3 - x)cos(π/3 + x).</p><p>Using cos(A-B)cos(A+B) = cos²A - sin²B with A = π/3 and B = x:</p><p>cos(π/3 - x)cos(π/3 + x) = cos²(π/3) - sin²(x) = 1/4 - sin²(x) = cos²(x) - 3/4</p><p><strong>Step 2:</strong> Substitute into original equation.</p><p>cos(x)[cos²(x) - 3/4] = 1/4</p><p>cos³(x) - (3/4)cos(x) = 1/4</p><p><strong>Step 3:</strong> Use cos(3x) = 4cos³(x) - 3cos(x).</p><p>Rearranging: 4cos³(x) - 3cos(x) = cos(3x)</p><p>So: cos³(x) - (3/4)cos(x) = (1/4)cos(3x)</p><p>Therefore: (1/4)cos(3x) = 1/4 ⟹ cos(3x) = 1</p><p><strong>Step 4:</strong> Solve cos(3x) = 1 in (0, 2π).</p><p>3x = 0, 2π, 4π, 6π, ...</p><p>x = 0, 2π/3, 4π/3, 2π</p><p>Valid solutions in (0, 2π): x = 2π/3, 4π/3</p><p><strong>Step 5:</strong> Sum the solutions.</p><p>2π/3 + 4π/3 = 6π/3 = 2π</p><p>∴ Answer: D</p>
Correct Answer: D