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If in a triangle $\mathrm{ABC} \mathrm{a} \cos ^2\left(\frac{\mathrm{C}}{2}\right)+\mathrm{c}^2 \cos ^2\left(\frac{\mathrm{A}}{2}\right)=\frac{3 \mathrm{~b}}{2}$, then the sides $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$
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are in A.P.
are in A.P.
If $a \cos ^2\left(\frac{\mathrm{C}}{2}\right)+\cos ^2\left(\frac{\mathrm{A}}{2}\right)=\frac{3 \mathrm{~b}}{2}$
$\mathrm{a}[\cos \mathrm{C}+1]+\mathrm{c}[\cos \mathrm{A}+1]=3 \mathrm{~b}$
$(\mathrm{a}+\mathrm{c})+(\mathrm{a} \cos \mathrm{C}+\mathrm{c} \cos \mathrm{B})=3 \mathrm{~b}$
$\mathrm{a}+\mathrm{c}+\mathrm{b}=3 \mathrm{~b}$ or $\mathrm{a}+\mathrm{c}=2 \mathrm{~b}$ or $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P.
$\mathrm{a}[\cos \mathrm{C}+1]+\mathrm{c}[\cos \mathrm{A}+1]=3 \mathrm{~b}$
$(\mathrm{a}+\mathrm{c})+(\mathrm{a} \cos \mathrm{C}+\mathrm{c} \cos \mathrm{B})=3 \mathrm{~b}$
$\mathrm{a}+\mathrm{c}+\mathrm{b}=3 \mathrm{~b}$ or $\mathrm{a}+\mathrm{c}=2 \mathrm{~b}$ or $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P.
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