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If in a triangle A B C ,cosA+2cosB+cosC=...

If in a triangle `A B C ,cosA+2cosB+cosC=2` prove that the sides of the triangle are in `AP`

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`cosA+2cosB+cosC = 2`
`=>cosA+cosC = 2(1-cosB)`
`=>2cos((A+C)/2)cos((A-C)/2) = 2(2sin^2(B/2))`
`=>2cos((pi-B)/2)cos((A-C)/2) = 2(2sin^2(B/2))`
`=>sin(B/2)cos((A-C)/2) = 2sin^2(B/2)`
`=>cos((A-C)/2) = 2sin(B/2)`
`=>cos(B/2)cos((A-C)/2) = 2sin(B/2)cos(B/2)`
`=>cos((pi-(A+C))/2)cos((A-C)/2) = sinB`
...
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