Question

please answer the attachment

Answer 1

sin A=cos²A

cos⁴A=sin²A=1-sinA
cos^(6) A = sin²A (1-sinA) = (1-sinA)²
=1-2sinA+1-sinA = 2-3sinA

{cos^(6)A}(acos^(6)A+bcos²A+c)=0
{2-3sinA}(2a-3asinA+bsinA+c)=0


Above is true when b=3 and 2a=-c

So, b+(c/a)+b=3-2+3=4