If A and B are acute angles such that sin A = | vedclass

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(A) Given: $\sin A = \sin^2 B$ and $2 \cos^2 A = 3 \cos^2 B$. Since $A$ and $B$ are acute,$\sin A, \sin B, \cos A, \cos B > 0$. Substitute $\cos^2 A =

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$\left(\frac{\pi}{6}, \frac{\pi}{4}\right)$

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(A) Given: $\sin A = \sin^2 B$ and $2 \cos^2 A = 3 \cos^2 B$. Since $A$ and $B$ are acute,$\sin A, \sin B, \cos A, \cos B > 0$. Substitute $\cos^2 A = 1 - \sin^2 A$ and $\cos^2 B = 1 - \sin^2 B$ into the second equation: $2(1 - \sin^2 A) = 3(1 - \sin^2 B)$. Substitute $\sin^2 B = \sin A$: $2 - 2 \sin^2 A = 3(1 - \sin A) = 3 - 3 \sin A$. $2 \sin^2 A - 3 \sin A + 1 = 0$. $(2 \sin A - 1)(\sin A - 1) = 0$. So,$\sin A = \frac{1}{2}$ or $\sin A = 1$. Since $A$ is an acute angle,$\sin A = 1$ implies $A = \frac{\pi}{2}$,which is not acute. Thus,$\sin A = \frac{1}{2}$,which gives $A = \frac{\pi}{6}$. Then $\sin^2 B = \sin A = \frac{1}{2}$,so $\sin B = \frac{1}{\sqrt{2}}$ (since $B$ is acute). Therefore,$B = \frac{\pi}{4}$. The solution is $(A, B) = \left(\frac{\pi}{6}, \frac{\pi}{4}\right)$.

If $A$ and $B$ are acute angles such that $\sin A = \sin^2 B$ and $2 \cos^2 A = 3 \cos^2 B$,then $(A, B) =$

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