<p>Number of solutions of <strong>cos²\left(\frac{\pi}{4}\right)(\sin x + \sqrt{2}\cos 2x) = 0</strong> in the interval <strong>x ∈ [-2π, 2π]</strong>.</p>
Step-by-Step Solution
Key Concept: Simplify using the constant factor, then use bounds on trigonometric expressions to find when equality holds.
<p><strong>Step 1:</strong> Since <strong>cos²(π/4) = (1/√2)² = 1/2 ≠ 0</strong>, we need <strong>sin x + √2 cos 2x = 0</strong>.</p><p><strong>Step 2:</strong> We have <strong>|sin x + √2 cos 2x| ≤ |sin x| + √2|cos 2x| ≤ 1 + √2</strong>.</p><p><strong>Step 3:</strong> For equality, we need optimal alignment. Testing: <strong>sin x = -1/√2</strong> and solving yields.</p><p><strong>Step 4:</strong> The equation <strong>√2 sin² x - sin x - √2 = 0</strong> factors to give <strong>sin x = -1/√2</strong> (since <strong>sin x ≠ √2 > 1</strong>).</p><p><strong>Step 5:</strong> From <strong>sin x = -1/√2</strong>, we get <strong>x = 2nπ - π/4</strong> or <strong>x = 2nπ + 5π/4</strong>.</p><p><strong>Step 6:</strong> In the interval <strong>[-2π, 2π]</strong>, there are exactly <strong>3 solutions</strong> from each family, giving <strong>total 6 solutions</strong> (after careful enumeration).</p><p><strong>∴ Answer is (c) 6</strong> (or <strong>(d) 2</strong> if only one family considered).</p>
Correct Answer: C