<p>If sec x cos 5x + 1 = 0, where 0 < x < 2π, then x = ?</p>
Step-by-Step Solution
Key Concept: Rewrite sec x cos 5x = -1 as a cosine equation and use periodicity.
<p><strong>Step 1:</strong> Given sec x cos 5x + 1 = 0</p><p><strong>Step 2:</strong> This gives sec x cos 5x = -1, or cos 5x/cos x = -1</p><p><strong>Step 3:</strong> Therefore cos 5x = -cos x = cos(π - x)</p><p><strong>Step 4:</strong> So 5x = ±(π - x) + 2nπ</p><p><strong>Step 5:</strong> Taking 5x = π - x + 2nπ gives 6x = π + 2nπ, so x = π/6 + nπ/3. In (0, 2π): x = π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6</p><p>∴ Answer is (a).</p>
Correct Answer: a