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: Simplify the RHS using the identity 1 + cos x = 2cos²(x/2) and tan(x/2) = sin(x/2)/cos(x/2), then convert both sides to the form R·sin(2x + φ) = constant to find solvable conditions.
<p><strong>Step 1:</strong> Simplify RHS using half-angle identities.</p><p>RHS = cos 2x + 2tan(x/2)·(1 + cos x)</p><p>Since 1 + cos x = 2cos²(x/2) and tan(x/2) = sin(x/2)/cos(x/2):</p><p>RHS = cos 2x + 2·[sin(x/2)/cos(x/2)]·2cos²(x/2)</p><p>RHS = cos 2x + 4sin(x/2)cos(x/2)</p><p>RHS = cos 2x + 2sin x</p><p><strong>Step 2:</strong> Rewrite the equation.</p><p>√3 sin 2x = cos 2x + 2sin x</p><p>√3 sin 2x - cos 2x = 2sin x</p><p><strong>Step 3:</strong> Convert to single trigonometric function.</p><p>2[√3/2 sin 2x - 1/2 cos 2x] = 2sin x</p><p>2sin(2x - π/6) = 2sin x</p><p>sin(2x - π/6) = sin x</p><p><strong>Step 4:</strong> Solve using sin A = sin B conditions.</p><p>Either 2x - π/6 = x + 2nπ ⟹ x = π/6 + 2nπ</p><p>Or 2x - π/6 = π - x + 2nπ ⟹ 3x = 7π/6 + 2nπ ⟹ x = 7π/18 + 2nπ/3</p><p>∴ Answer: B</p>
Correct Answer: B