Trigonometry - Finding value of expression

Question

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!

Answer 1

HINT:

$$\sin x=1-\sin^2x=\cos^2x$$

Squaring we get $$\sin^2x=(\cos^2x)^2\iff 1-c^2=c^4\iff c^4+c^2-1=0$$ where $c=\cos x$

Now establish that $c^4+c^2-1$ is a factor of $$c^{12} + 3c^{10} + 3c^8 + 6c^6 + 2c^4 + c^2 -2$$

Answer 2

$\sin x+\sin ^2x=1$, 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?