In triangle ABC,if a cos^2 frac{C}{2} + c cos | vedclass

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(C) Given $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get:$\frac{a

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$2: 1$

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(C) Given $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get:$\frac{a(1 + \cos C) + c(1 + \cos A)}{2} = \frac{3b}{2}$.$a + a \cos C + c + c \cos A = 3b$.By the projection rule in $	riangle ABC$,$a \cos C + c \cos A = b$.Substituting this into the equation:$a + c + b = 3b$.$a + c = 2b$.Therefore,$\frac{a+c}{b} = \frac{2}{1}$,which means $a+c : b = 2 : 1$.

In $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then $a+c : b =$

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