Trigonometry & Inverse Trigonometry
Trigonometric equations
Grade 11
Question:
<p>For the equation \(\sqrt{3}\sin 2x = \cos 2x + 2\tan\dfrac{x}{2}(1 + \cos x)\), which of the following holds good?</p>
<p>(a) The number of solutions of the equation in \([0, 2\pi]\) is 4</p>
<p>(b) The number of solutions of the equation in \([0, 2\pi]\) is 3</p>
<p>(c) If \(\alpha\) is the smallest positive root of the equation then \(\dfrac{\tan 2\alpha + 2\cos 2\alpha}{\cot \alpha - \sin 3\alpha} = 2 + \sqrt{3}\)</p>
<p>(d) If \(\alpha\) is the smallest positive root of the equation then \(\dfrac{\tan 2\alpha + 2\cos 4\alpha}{\cot \alpha + \sin 3\alpha} = 2 - \sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: Convert the equation using the identity tan(x/2) = sin(x)/(1+cos(x)) and express everything in terms of sin 2x and cos 2x to identify the solution set. The term 2tan(x/2)(1+cos x) simplifies to 2sin(x), which is the key to linearizing the trigonometric equation.
<p><strong>Step 1:</strong> Simplify the RHS term: 2tan(x/2)(1+cos x) = 2·sin(x)/(1+cos x)·(1+cos x) = 2sin x</p><p><strong>Step 2:</strong> Rewrite the equation: √3 sin 2x = cos 2x + 2sin x</p><p><strong>Step 3:</strong> Use sin 2x = 2sin x cos x: √3·2sin x cos x = cos 2x + 2sin x</p><p><strong>Step 4:</strong> Rearrange: 2√3 sin x cos x - 2sin x = cos 2x</p><p><strong>Step 5:</strong> Factor: 2sin x(√3 cos x - 1) = cos 2x</p><p><strong>Step 6:</strong> Expand cos 2x = 1 - 2sin²x and simplify to get: 2√3 sin x cos x - 2sin x - cos 2x = 0</p><p><strong>Step 7:</strong> This simplifies to sin(2x - π/6) = 1/2, giving 2x - π/6 = π/6 + 2nπ or 5π/6 + 2nπ</p><p><strong>Step 8:</strong> Solve for x: x = π/6 + nπ or x = π/2 + nπ, with domain restriction x ≠ (2n+1)π</p><p>∴ Answer: B</p>
Correct Answer: B