If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$

I write it step by step:

With $\sin A+\sin^2 A=1$ we have $\sin A=1-\sin^2 A=\cos^2A$ so \begin{eqnarray} && a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0\\ && a(\cos^2A)^6+b(\cos^2A)^4+c(\cos^2A)^3-1=0\\ && a\sin^6A+b\sin^4A+c\sin^3A-1=0\\ && a(1-\cos^2A)^3+b(1-\cos^2A)^2+c\sin A(1-\cos^2A)-1=0\\ && a(1-\sin A)^3+b(1-\sin A)^2+c\sin A(1-\sin A)-1=0\\ && a(1-3\sin A+3\sin^2A-\sin^3A+b(1-2\sin A+\sin^2A)+c(\sin A-\sin^2 A)-1=0\\ && a-3a\sin A+3a\sin^2A-a\sin^3A+b-2b\sin A+b\sin^2A+c\sin A-c\sin^2 A-1=0\\ && a-3a\sin A+3a\sin^2A-a\sin^3A+b-2b\sin A+b\sin^2A+c\sin A-c\sin^2 A-1=0\\ && a-3a\sin A+3a(1-\cos^2A)-a\sin A(1-\cos^2A)+b-2b\sin A+b(1-\cos^2A)+c\sin A-c(1-\cos^2A)-1=0\\ && a-3a\sin A+3a(1-\sin A)-a\sin A(1-\sin A)+b-2b\sin A+b(1-\sin A)+c\sin A-c(1-\sin A)-1=0\\ && a-3a\sin A+3a-3a\sin A-a\sin A+a\sin^2A+b-2b\sin A+b-b\sin A+c\sin A-c+c\sin A-1=0\\ && a-3a\sin A+3a-3a\sin A-a\sin A+a-a\sin A+b-2b\sin A+b-b\sin A+c\sin A-c+c\sin A-1=0\\ && (a+3a+a+b+b-c-1)+(-3a-3a-a-a-2b-b+c+c)\sin A=0\\ && (5a+2b-c-1)+(-8a-3b+2c)\sin A=0 \end{eqnarray}

$$\sin A= \cos ^2A→\sin^2A=\cos^4A→1-\cos^2A=\cos^4A\\ (\cos^2A)^2+(\cos^2A)-1=0→\cos^2A=\frac{-1+ \sqrt{5}}{2}$$

So,

$$a\left(\frac{-1+ \sqrt{5}}{2}\right)^6+b\left(\frac{-1+ \sqrt{5}}{2}\right)^4+c\left(\frac{-1+ \sqrt{5}}{2}\right)^3=1\\ a(9-4\sqrt{5})+b\left(\frac{7-3\sqrt{5}}{2}\right)+c(\sqrt{5}-2)=1\\ (9a+7b/2-2c)+(-4a-3b/2+c)\sqrt{5}=1$$

Maybe the question want:

$$9a+\frac{7b}{2}-2c=1→18a+7b-4c=2\\ -4a-\frac{3b}{2}+c=0→8a+3b=2c$$

But on that case we get:

$$b=2-2a\\ c=a+3$$

And then

$$2b+\frac{c}{a}=5-4a+\frac{3}{a}$$

and the result depends on $a$. Otherwise I don't have another guess.