<s>[INST] Find $a+b +c$, if $\sin{x}+\sin^2{x}=1$ and $\cos^{12}{x}+a\cos^{10}{x}+b\cos^8{x}+c\cos^6{x}=1$ There is my problem :
Find $a+b +c$,
if $$\sin{x}+\sin^2{x}=1$$ and $$\cos^{12}{x}+a\cos^{10}{x}+b\cos^8{x}+c\cos^6{x}=1$$
I'm sorry, I can't solve this problem but I really want to know the solution.
I know that $\cos^2{x}=\sin{x}$, but I can't find $a+b+c$.
Attempt
I used substitute $t=\sin(x)$, and number $1=t^2+t$ put on the left. Then I divided by $t$ as long as I can, then I got polynomial with degree $3$, but I can't conclude what is $a+b+c$.
 [/INST] $\sin^2 x+\sin x-1=0$ gives $\sin x = \frac{-1\pm\sqrt{5}}{2}$, so that
$$\cos^2 x = 1-\sin^2 x = 1-\frac{3\pm\sqrt{5}}{2} = \frac{-1\pm\sqrt{5}}{2}.$$
Then you want to solve
$$(\cos^2 x)^6+a(\cos^2 x)^5 + b(\cos^2 x)^4 + c(\cos^2 x)^3 = 1.$$
After some algebra, this reduces to (for the positive sign on $\sqrt{5}$)
\begin{gather*}
-\frac{11 }{2}a+\frac{7 }{2}b-2 c+9=1 \\
4 - \frac{5}{2}a + \frac{3}{2} b - c = 0
\end{gather*}
which has the solution $a=b$, $c=4-b$. So the best you can do is $b+c=4$.
Taking the negative sign on $\sqrt{5}$ produces the same result.
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