<p>Solve <br>\(\sin[2\cos^{-1}\{\cot(2\tan^{-1}x)\}] = 0\); express \(x\) in the form \(a \pm b\).</p>
Step-by-Step Solution
Key Concept: Decompose the equation from outside to inside: sin(...) = 0 implies the inner expression equals 0 or π, then systematically simplify inverse trigonometric compositions using algebraic identities and substitution techniques.
<p><strong>Step 1: Apply sin(θ) = 0 condition</strong></p><p>sin[2cos⁻¹{cot(2tan⁻¹ x)}] = 0</p><p>This requires: 2cos⁻¹{cot(2tan⁻¹ x)} = nπ, where n ∈ ℤ</p><p>Since cos⁻¹ has range [0, π], we have 2cos⁻¹{...} ∈ [0, 2π]</p><p>Thus: 2cos⁻¹{cot(2tan⁻¹ x)} = 0 or π</p><p></p><p><strong>Step 2: Solve Case 1 — 2cos⁻¹{cot(2tan⁻¹ x)} = 0</strong></p><p>cos⁻¹{cot(2tan⁻¹ x)} = 0</p><p>cot(2tan⁻¹ x) = cos(0) = 1</p><p></p><p><strong>Step 3: Simplify 2tan⁻¹ x</strong></p><p>Let tan⁻¹ x = α, so tan α = x</p><p>tan(2α) = (2tan α)/(1 - tan² α) = 2x/(1 - x²)</p><p>cot(2α) = (1 - x²)/(2x) = 1</p><p></p><p><strong>Step 4: Solve (1 - x²)/(2x) = 1</strong></p><p>1 - x² = 2x</p><p>x² + 2x - 1 = 0</p><p>x = (-2 ± √(4 + 4))/2 = (-2 ± 2√2)/2 = -1 ± √2</p><p></p><p><strong>Step 5: Solve Case 2 — 2cos⁻¹{cot(2tan⁻¹ x)} = π</strong></p><p>cos⁻¹{cot(2tan⁻¹ x)} = π/2</p><p>cot(2tan⁻¹ x) = cos(π/2) = 0</p><p></p><p><strong>Step 6: Simplify cot(2tan⁻¹ x) = 0</strong></p><p>cot(2α) = (1 - x²)/(2x) = 0</p><p>1 - x² = 0</p><p>x² = 1</p><p>x = ±1</p><p></p><p><strong>Step 7: Verification</strong></p><p>For x = 1: cot(2·45°) = cot(90°) = 0 ✓</p><p>For x = -1: cot(2·(-45°)) = cot(-90°) = 0 ✓</p><p>For x = -1 ± √2: cot(2tan⁻¹ x) = 1 ✓</p><p>For x = 1 ± √2: cot(2tan⁻¹ x) = 1 ✓</p><p></p><p><strong>∴ Answer:</strong> $x = 1, -1, -1+\sqrt{2}, -1-\sqrt{2}, 1+\sqrt{2}, 1-\sqrt{2}$</p>
Correct Answer: \(x = 1, -1, -1+\sqrt{2}, -1-\sqrt{2}, 1+\sqrt{2}, 1-\sqrt{2}\)