<p>If \(\left(\cos^2 x + \dfrac{1}{\cos^2 x}\right)(1 + \tan^2 2y)(3 + \sin 3z) = 4\), then</p>
<p>(a) x may be a multiple of \(\pi\)</p>
<p>(b) x can not be an even multiple of \(\pi\)</p>
<p>(c) z can be a multiple of \(\pi\)</p>
<p>(d) y can be multiple of \(\dfrac{\pi}{2}\)</p>
Step-by-Step Solution
Key Concept: Each factor must equal its minimum value simultaneously: (cos²x + 1/cos²x) ≥ 2 by AM-GM, (1 + tan²2y) = sec²2y ≥ 1, and (3 + sin3z) ∈ [2,4]. Since their product equals 4, we need cos²x + 1/cos²x = 2, and the other factors optimally constrained.
<p><strong>Step 1:</strong> Analyze each factor's range using AM-GM inequality.</p><p>For the first factor: cos²x + 1/cos²x ≥ 2√(cos²x · 1/cos²x) = 2, with equality when cos²x = 1, i.e., cos x = ±1, so x = nπ.</p><p><strong>Step 2:</strong> The second factor: (1 + tan²2y) = sec²2y ≥ 1, with minimum when tan 2y = 0, i.e., 2y = nπ, so y = nπ/2.</p><p><strong>Step 3:</strong> The third factor: (3 + sin 3z) where sin 3z ∈ [-1,1], so (3 + sin 3z) ∈ [2,4].</p><p><strong>Step 4:</strong> For the product to equal 4: We need (cos²x + 1/cos²x) = 2, (1 + tan²2y) = 1, and (3 + sin 3z) = 2.</p><p>This gives: cos²x = 1 → x = nπ; tan 2y = 0 → y = nπ/2; sin 3z = -1 → 3z = -π/2 + 2πk → z = -π/6 + 2πk/3.</p><p>∴ Answer: AC</p>
Correct Answer: AC