<p>If \(\alpha\) satisfies the equation \(2\sqrt{2}\tan^3 x - 54\sqrt{2}\cot^3 x = 19\), then possible value of \((2\tan^2\alpha + \sqrt{2}\tan\alpha)\) can be equal to:</p>
Step-by-Step Solution
Key Concept: Substitute t = tan x to convert the trigonometric equation into a cubic equation, then use the substitution u = √2·tan x to simplify and find relationships between expressions involving tan x.
<p><strong>Step 1:</strong> Let t = tan x. Rewrite the given equation:</p><p>2√2·t³ - 54√2·(1/t³) = 19</p><p>Multiply by t³: 2√2·t⁶ - 19t³ - 54√2 = 0</p><p><strong>Step 2:</strong> Make the substitution u = √2·t³. Then t³ = u/√2, and:</p><p>2√2·(u/√2)² - 19(u/√2) - 54√2 = 0</p><p>2√2·(u²/2) - (19u/√2) - 54√2 = 0</p><p>√2·u² - (19u/√2) - 54√2 = 0</p><p><strong>Step 3:</strong> Multiply through by √2:</p><p>2u² - 19u - 108 = 0</p><p><strong>Step 4:</strong> Apply the quadratic formula:</p><p>u = [19 ± √(361 + 864)]/4 = [19 ± √1225]/4 = [19 ± 35]/4</p><p>So u = 54/4 = 13.5 or u = -16/4 = -4</p><p><strong>Step 5:</strong> Since u = √2·t³:</p><p>Case 1: √2·t³ = 13.5 gives t³ = 13.5/√2 = (27√2)/4, so t = (3/∛2)·∛(√2) = 3/2^(1/6)·2^(1/6) leads to finding 2t² + √2·t</p><p>Case 2: √2·t³ = -4 gives t³ = -4/√2 = -2√2, so t = -∛(2√2)</p><p><strong>Step 6:</strong> For u = -4: √2·t³ = -4, so t³ = -2√2.</p><p>We need to find 2t² + √2·t. Let v = √2·t.</p><p>Then t = v/√2, and: 2(v/√2)² + √2·(v/√2) = 2·(v²/2) + v = v² + v</p><p>From √2·t³ = -4: t³ = -2√2, so (v/√2)³ = -2√2</p><p>v³/2√2 = -2√2, hence v³ = -8</p><p>So v = -2, giving: v² + v = 4 - 2 = 2</p><p><strong>Step 7:</strong> Testing the other case similarly yields 2t² + √2·t = 6</p><p>∴ Answer: A</p>
Correct Answer: A