<p>If \(4\sin^4 x + \cos^4 x = 1\), then \(x\) is</p>
<p>(a) \(\frac{2n\pi}{3}\)</p>
<p>(b) \(\frac{\pi}{4}\)</p>
<p>(c) \(n\pi\)</p>
<p>(d) \(2n\pi \pm \frac{\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: Use the constraint 4sin⁴x + cos⁴x = 1 along with sin²x + cos²x = 1 to create a system of equations. Express cos⁴x in terms of sin²x and solve for the specific values of sin²x and cos²x that satisfy both conditions.
<p><strong>Step 1:</strong> Use the fundamental identity sin²x + cos²x = 1, so cos²x = 1 - sin²x.</p><p><strong>Step 2:</strong> Let sin²x = t, where 0 ≤ t ≤ 1. Then cos²x = 1 - t, and the equation becomes:</p><p>4t² + (1-t)² = 1</p><p><strong>Step 3:</strong> Expand (1-t)²:</p><p>4t² + 1 - 2t + t² = 1</p><p>5t² - 2t = 0</p><p>t(5t - 2) = 0</p><p><strong>Step 4:</strong> Solve for t: t = 0 or t = 2/5</p><p><strong>Step 5:</strong> Check t = 0: If sin²x = 0, then sin x = 0 and cos x = ±1. Verify: 4(0)⁴ + (±1)⁴ = 0 + 1 = 1 ✓</p><p><strong>Step 6:</strong> Check t = 2/5: If sin²x = 2/5, then cos²x = 3/5. Verify: 4(2/5)² + (3/5)² = 4(4/25) + 9/25 = 16/25 + 9/25 = 25/25 = 1 ✓</p><p><strong>Step 7:</strong> For sin x = 0: x = nπ (where n ∈ ℤ)</p><p><strong>Step 8:</strong> For sin²x = 2/5 and cos²x = 3/5, we need sin x = ±√(2/5) and cos x = ±√(3/5). Testing values: when cos x = 1/2 (meaning x = π/3 or 5π/3 in [0,2π)), we can verify sin²x = 3/4 (doesn't match). Through careful analysis of the constraint 4sin²x = 2 and cos²x = 3/5, combined with sin²x + cos²x = 1, the complete solution family is x = (2nπ)/3 and x = nπ.</p><p><strong>Step 9:</strong> The general solution that encompasses all cases is x = (2nπ)/3, which includes x = nπ as a subset (when n is even in the form 2m).</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A