Find the value of $\cos^{12}\theta + 3\cos^{10}\theta + 3\cos^{8}\theta + \cos^6\theta + 2\cos^4\theta + 2\cos^2\theta - 2$ We are given that $\sin\theta + \sin^3\theta + \sin^2\theta = 1$
Find the value of $\cos^{12}\theta + 3\cos^{10}\theta + 3\cos^{8}\theta + \cos^6\theta + 2\cos^4\theta + 2\cos^2\theta - 2$
Now, I was able to establish the following from the first equation: 
$\sin\theta + \sin^3\theta + \sin^2\theta = 1 = \sin^2\theta + \cos^2\theta \implies \sin\theta + \sin^3\theta = \cos^2\theta$
The next obvious step was to simplify the second expression.  I let $\cos^2\theta = x$:
$f(x) = x^6 + 3x^5 + 3x^4 + x^3 + 2x^2 + 2x - 2$
$f(-1) = 0 \implies f(x) = (x + 1)(x^5 + 2x^4 + x^3 + 2x - 2)$
I was stuck after this. 

Observe that the powers of $\cos \theta$ are all even, suggesting that we should use the conversion $\cos^2 \theta = 1 - \sin ^2 \theta$. For simplicity, let $x = \sin \theta$. We are given that
$$(x + x^2 + x^3) = 1$$
and want to find
$$(1-x^2)^6 + 3(1-x^2)^5 + 3(1-x^2)^4 + (1-x^2)^3 + 2(1-x^2)^2 + 2(1-x^2) - 2 $$
By long division, we could factor out $x^3 + x^2 + x -1 $ to get $214x^2 + 20x - 68$, but then it's not clear what to do after that.

You could use the fact that $x^3 + x^2 + x - 1 = 0 $ has 1 real root; but I doubt that is how they want you to proceed.
