Trigonometry - Finding value of expression Question: Given $\sin x + \sin^2 x = 1$ then find the value of $$\cos^{12} x + 3\cos^{10} x + 3\cos^8 x + 6\cos^6 x + 2\cos^4 x + \cos^2 x -2$$
I have no idea where to start on this question. Please help me!

$\sin x+\sin ^2x=1$, let's give $\sin x$ another name, $\alpha$.
$\alpha+\alpha ^2 =1$.
$\alpha^2 + \alpha -1=0$.
That gives us either
$\alpha_1=\frac{-1+\sqrt{5}}{2}$ or $\alpha_2=\frac{-1-\sqrt{5}}{2}$.
Notice that $\sin(x) = \frac{-1-\sqrt{5}}{2}$ can't be true. So the only valid solution is $\alpha_1$.
Given that $\sin^2x=(\frac{-1+\sqrt{5}}{2})^2$, this means that $\cos^2x=1-(\frac{-1+\sqrt{5}}{2})^2$
Can you take it from here?
