If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$ If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$.
Find the value of $2b + \dfrac {c}{a}$.
My Attempt:
$$\sin A+\sin^2 A=1$$
$$\sin A + 1 - \cos^2 A=1$$
$$\sin A=\cos^2 A$$
Now,
$$a(\cos^2 A)^{6}+b(\cos^2 A)^{4}+c(\cos^2 A)^{3}=1$$
$$a\sin^6 A+ b\sin^4 A+c\sin^3 A=1$$
How do I proceed further?
 A: 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}
A: $$\sin A= \cos ^2A→\sin^2A=\cos^4A→1-\cos^2A=\cos^4A\\
(\cos^2A)^2+(\cos^2A)-1=0→\cos^2A=\frac{-1+ \sqrt{5}}{2}$$
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.
A: @MyGlasses .. In the third line from below you forgot that there is $(-a)$ twice in your attempt to prove the required identity, resulting in a mistake in your answer. 
A: HINT:
$$\cos^4A=(\cos^2A)^2=\cdots=1-\cos^2A\iff\cos^4A+\cos^2A-1=0$$
Divide $a\cos^{12}A+b\cos^8A+c\cos^8A-1$ by $\cos^4A+\cos^2A-1$
