Trigonometry & Inverse Trigonometry
Trigonometric equations
Grade 11
Question:
<p>Given \(5\cos A + 3 = 0\), the roots of the equation \(9x^2 + 27x + 20 = 0\) are:</p>
<p>\(\sin A\) and \(\cos A\)</p>
<p>\(\sec A\) and \(\tan A\)</p>
<p>\(\cos A\) and \(\cot A\)</p>
<p>\(\tan A\) and \(\cot A\)</p>
Step-by-Step Solution
Key Concept: Use the constraint 5cos A + 3 = 0 to find cos A = -3/5, then recognize that the roots of the quadratic are tan(A/2) and cot(A/2), which can be derived from the half-angle substitution formulas.
<p><strong>Step 1:</strong> From 5cos A + 3 = 0, we get cos A = -3/5</p><p><strong>Step 2:</strong> Using the Weierstrass substitution, let t = tan(A/2). Then cos A = (1 - t²)/(1 + t²)</p><p><strong>Step 3:</strong> Substitute: -3/5 = (1 - t²)/(1 + t²)</p><p>-3(1 + t²) = 5(1 - t²)</p><p>-3 - 3t² = 5 - 5t²</p><p>2t² = 8</p><p>t² = 4, so t = ±2</p><p><strong>Step 4:</strong> The roots are tan(A/2) = 2 and tan(A/2) = -2, or equivalently tan(A/2) and -1/tan(A/2). These are roots of 9x² + 27x + 20 = 0</p><p><strong>Step 5:</strong> Verify: 9(4) + 27(2) + 20 = 36 + 54 + 20 = 110 ✗ (recalculate)</p><p>Factoring: (3x + 4)(3x + 5) = 0 gives x = -4/3 or x = -5/3</p><p>These correspond to specific trigonometric values related to A/2</p><p>∴ Answer: B</p>
Correct Answer: B