Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>The value of <span>x</span> in <span>\left(0, \frac{π}{2}\right)</span> satisfying the equation <span>\frac{\sqrt{5}-1}{4} \cdot \frac{1}{\sin x} + \frac{\sqrt{10+2\sqrt{5}}}{4} \cdot \frac{1}{\cos x} = 2</span> is</p>
<p>(a) <span>\frac{π}{10}</span></p>
<p>(b) <span>\frac{3π}{10}</span></p>
Step-by-Step Solution
Key Concept: Express the trigonometric combination in the form sin(x + α) or cos(x + α) to simplify the equation, then solve the resulting simple equations.
<p><strong>Step 1:</strong> Simplify the equation: <span>\frac{\sqrt{5}-1}{4} \csc x + \frac{\sqrt{10+2\sqrt{5}}}{4} \sec x = 2</span></p><p><strong>Step 2:</strong> Multiply through and rearrange: <span>(\sqrt{5}-1) \cos x + \sqrt{10+2\sqrt{5}} \sin x = 8 \sin x \cos x = 4 \sin 2x</span></p><p><strong>Step 3:</strong> This can be written as: <span>\sin\left(x + \frac{π}{10}\right) = \sin 2x</span></p><p><strong>Step 4:</strong> Case 1: <span>x + \frac{π}{10} = 2x</span>, giving <span>x = \frac{π}{10}</span></p><p><strong>Step 5:</strong> Case 2: <span>\sin\left(x + \frac{π}{10}\right) = \sin(π - 2x)</span>, giving <span>x + \frac{π}{10} = π - 2x</span>, so <span>3x = \frac{9π}{10}</span></p><p>∴ Answer is (a) <span>\frac{π}{10}</span></p>
Correct Answer: a