<p>For a constant <em>k</em>, the two roots of the quadratic equation \(3x^2 - x + k = 0\) are \(\sin\theta\) and \(\cos\theta\). The value of \(54(\sin^3\theta + \cos 3\theta)\) is:</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to relate sin θ and cos θ as roots, then employ the identity sin³θ + cos³θ = (sin θ + cos θ)³ - 3sin θ cos θ(sin θ + cos θ) to express the cubic sum in terms of elementary symmetric functions.
<p><strong>Step 1: Apply Vieta's Formulas</strong></p><p>If sin θ and cos θ are roots of 3x² - x + k = 0, then:</p><p>• Sum of roots: sin θ + cos θ = 1/3</p><p>• Product of roots: sin θ · cos θ = k/3</p><p><strong>Step 2: Use Fundamental Trigonometric Identity</strong></p><p>Since sin²θ + cos²θ = 1:</p><p>(sin θ + cos θ)² - 2sin θ cos θ = 1</p><p>(1/3)² - 2(k/3) = 1</p><p>1/9 - 2k/3 = 1</p><p>2k/3 = -8/9</p><p>k = -4/3, so sin θ cos θ = -4/9</p><p><strong>Step 3: Calculate sin³θ + cos³θ</strong></p><p>sin³θ + cos³θ = (sin θ + cos θ)³ - 3sin θ cos θ(sin θ + cos θ)</p><p>= (1/3)³ - 3(-4/9)(1/3)</p><p>= 1/27 + 4/9</p><p>= 1/27 + 12/27 = 13/27</p><p><strong>Step 4: Compute 54(sin³θ + cos³θ)</strong></p><p>54 × 13/27 = 2 × 13 = 26</p><p>∴ Answer: <strong>C</strong></p>
Correct Answer: C