<s>[INST] 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. 
 [/INST] Use that $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ and $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.$$
Then replace $e^{ix}$ by $z$ and clear denominators. The first condition becomes an equality of the form $f(z)=0$, the second a polynomial $g(z)$. Use long division to get $g=fq+r$. The value of $f$ will be the same as that of $r$. If you are lucky $r=0$. If not, you will have to solve $f(z)=0$ and evaluate $r(z)$.
I did the computations ... not pretty. $r\neq0$.
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