If in a \[\Delta ABC\] \[a\,{\cos ^2}\left( {{C \over 2}} \right) + c\,{\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2},\] then the sides a,b and c :
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Official solution
If \[a\,{\cos ^2}\left( {{C \over 2}} \right) + c\,{\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2}\]
\[a\left[ {\cos C + 1} \right] + c\left[ {\cos A + 1} \right] = 3b\]
\[\left( {a + c} \right) + \left( {a\cos C + c\cos \,B} \right) = 3b\]
a+c+b=3b or a+c=2b
or a,b,c are in A.P.