If A and B are acute angles such that sin A = sin2 B and 2 cos2 A = 3 cos2 B, then (A, B) =
\(\left(\dfrac{\pi}{6}, \dfrac{\pi}{4}\right)\)
Two equations in sin A and sin B can be reduced to one equation in a single variable by using \(\cos^{2}\theta = 1 - \sin^{2}\theta\). Because A and B are acute, all sines and cosines are positive, so each value found fixes the angle uniquely.
Step 1 — Let \(t = \sin^{2} B\). Then \(\sin A = t\), so \(\cos^{2} A = 1 - t^{2}\), and \(\cos^{2} B = 1 - t\).
Step 2 — The second condition becomes \(2(1 - t^{2}) = 3(1 - t)\), i.e. \(2t^{2} - 3t + 1 = 0\), so \((2t - 1)(t - 1) = 0\).
Step 3 — \(t = 1\) would give \(B = \dfrac{\pi}{2}\), which is not acute, so \(t = \dfrac{1}{2}\). Then \(\sin^{2} B = \dfrac{1}{2} \Rightarrow B = \dfrac{\pi}{4}\) and \(\sin A = \dfrac{1}{2} \Rightarrow A = \dfrac{\pi}{6}\).
Check: \(2\cos^{2}\dfrac{\pi}{6} = \dfrac{3}{2} = 3\cos^{2}\dfrac{\pi}{4}\). The pair (π/6, π/6) gives sin A = 1/2 but sin2 B = 1/4; (π/4, π/6) gives sin A = 1/√2 against 1/4; (π/4, π/4) gives 1/√2 against 1/2, so the first condition fails in all three.
\(\left(\dfrac{\pi}{6}, \dfrac{\pi}{4}\right)\) option is correct.
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