<p>Given \(1 + \sin^4 x = \cos^2 3x\), find the number of solutions for \(x \in \left[-\dfrac{5\pi}{2}, \dfrac{5\pi}{2}\right]\).</p>
Step-by-Step Solution
Key Concept: Recognize that 1 + sin⁴x ≥ 1 while cos²3x ≤ 1, so equality requires both sides equal 1. This means sin⁴x = 0 AND cos²3x = 1 simultaneously.
<p><strong>Step 1:</strong> Analyze the range constraints.</p><p>Since sin⁴x ≥ 0 for all x, we have: 1 + sin⁴x ≥ 1</p><p>Since cos²3x ≤ 1 for all x, we have: cos²3x ≤ 1</p><p><strong>Step 2:</strong> For equality to hold, both sides must equal 1.</p><p>1 + sin⁴x = 1 ⟹ sin⁴x = 0 ⟹ sin x = 0</p><p>cos²3x = 1 ⟹ cos 3x = ±1 ⟹ 3x = nπ ⟹ x = nπ/3</p><p><strong>Step 3:</strong> Find intersection of both conditions.</p><p>sin x = 0 ⟹ x = kπ (where k ∈ ℤ)</p><p>We need x = kπ to also equal nπ/3, which means x must be a multiple of π.</p><p><strong>Step 4:</strong> Count solutions in [-5π/2, 5π/2].</p><p>x = kπ where -5π/2 ≤ kπ ≤ 5π/2</p><p>-5/2 ≤ k ≤ 5/2</p><p>k ∈ {-2, -1, 0, 1, 2}</p><p>Solutions: x ∈ {-2π, -π, 0, π, 2π}</p><p>∴ Answer: <strong>5</strong></p>
Correct Answer: 5