<p>The number of solutions of the equation \(\tan x + \sec x = 2\cos x\) lying in the interval \([0, 2\pi]\) is</p>
Step-by-Step Solution
Key Concept: Rewrite the equation using tan x = sin x/cos x and sec x = 1/cos x, then multiply through by cos x (carefully tracking domain restrictions where cos x ≠ 0) to get a polynomial equation in sin x. The key is recognizing that cos x = 0 is not a solution and must be excluded from the domain.
<p><strong>Step 1:</strong> Start with tan x + sec x = 2cos x. Rewrite using tan x = sin x/cos x and sec x = 1/cos x:</p><p>sin x/cos x + 1/cos x = 2cos x</p><p><strong>Step 2:</strong> Multiply both sides by cos x (valid since cos x ≠ 0, as it would make sec x undefined):</p><p>sin x + 1 = 2cos² x</p><p><strong>Step 3:</strong> Use cos² x = 1 - sin² x:</p><p>sin x + 1 = 2(1 - sin² x)</p><p>sin x + 1 = 2 - 2sin² x</p><p>2sin² x + sin x - 1 = 0</p><p><strong>Step 4:</strong> Factor the quadratic in sin x:</p><p>(2sin x - 1)(sin x + 1) = 0</p><p><strong>Step 5:</strong> Solve each factor:</p><p>• 2sin x - 1 = 0 ⟹ sin x = 1/2 ⟹ x = π/6, 5π/6 in [0, 2π]</p><p>• sin x + 1 = 0 ⟹ sin x = -1 ⟹ x = 3π/2 in [0, 2π]</p><p><strong>Step 6:</strong> Verify each solution in the original equation:</p><p>• x = π/6: tan(π/6) + sec(π/6) = 1/√3 + 2/√3 = 3/√3 = √3; 2cos(π/6) = 2(√3/2) = √3 ✓</p><p>• x = 5π/6: tan(5π/6) + sec(5π/6) = -1/√3 - 2/√3 = -3/√3 = -√3; 2cos(5π/6) = 2(-√3/2) = -√3 ✓</p><p>• x = 3π/2: cos(3π/2) = 0, making sec(3π/2) undefined ✗</p><p>∴ Answer: <strong>2 solutions</strong></p>
Correct Answer: C