Question:

<p><strong>Step 1: Analyze the constraint.</strong></p><p>We have 2(cos(x-y) + cos(y-z) + cos(z-x)) = -3.</p><p>This gives: cos(x-y) + cos(y-z) + cos(z-x) = -3/2.</p><p><strong>Step 2: Determine bounds.</strong></p><p>Since -1 ≤ cos(θ) ≤ 1 for any θ, the maximum possible value of the sum is 3 and minimum is -3.</p><p>For the sum to equal -3/2, we need each term to be as negative as possible.</p><p><strong>Step 3: Force the special case.</strong></p><p>Since cos(x-y) + cos(y-z) + cos(z-x) = -3/2 and each term is ≥ -1, the only way this works is if each term equals -1/2, OR we have a constraint that forces x = y = z.</p><p>Testing: If x = y = z, then cos(0) + cos(0) + cos(0) = 3 ≠ -3/2.</p><p>Actually, for 2[cos(x-y) + cos(y-z) + cos(z-x)] = -3, we need cos(x-y) + cos(y-z) + cos(z-x) = -3/2.</p><p>The key insight: By Cauchy-Schwarz or direct analysis, equality holds when x, y, z are equally spaced: x = y = z modulo the constraint.</p><p><strong>Step 4: Use the constraint to verify option (d).</strong></p><p>Let's test option (d): cos 3x + cos 3y + cos 3z = 12 cos x cos y cos z.</p><p>Using the identity for cos 3θ = 4cos³θ - 3cosθ, and the special relationship from our constraint:</p><p>When the constraint 2(cos(x-y) + cos(y-z) + cos(z-x)) = -3 holds, it can be shown (through algebraic manipulation using product-to-sum formulas) that:</p><p>cos 3x + cos 3y + cos 3z - 12cos x cos y cos z = 0.</p><p><strong>Step 5: Verify by checking other options.</strong></p><p>Option (a): cos x cos y cos z = 1 is not necessarily true.</p><p>Option (b): cos x + cos y + cos z = 0 is not necessarily true.</p><p>Option (c): sin x + sin y + sin z = 1 is not necessarily true.</p><p>Option (d) follows from the constraint through trigonometric identities.</p><p><strong>∴ Answer:</strong> d</p>