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 $a+b+c$.
 A: $\sin^2 x+\sin x-1=0$ 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.
A: With what you already know, $$\sin^6x+a\sin^5x+b\sin^4x+c\sin^3x-1=0\qquad \sin^2x+\sin x-1=0$$ Now let $X=\sin (x)$. You would like the polynomial $X^2+X-1$ to divide $X^6+aX^5+bX^4+cX^3-1$. How to make that happen?
There would need to be a 4th degree polynomial $v=X^4+v_3X^3+v_2X^2+v_1X+1$ as the quotient.
$$\left(X^2+X-1\right)\left(X^4+v_3X^3+v_2X^2+v_1X+1\right)=X^6+aX^5+bX^4+cX^3-1$$
You can deduce quickly that $v_1=1$ and $v_2=2$ after comparing linear and quadratic coefficients.
$$\left(X^2+X-1\right)\left(X^4+v_3X^3+2X^2+X+1\right)=X^6+aX^5+bX^4+cX^3-1$$
Comparing cubic, quartic, and quintic coefficients:
$$\begin{align}
1+2-v_3&=c\\
2+v_3-1&=b\\
v_3+1&=a
\end{align}$$
You could eliminate $v_3$ by adding the first equation to each of the other two:
$$\begin{align}
4&=b+c\\
4&=a+c
\end{align}$$
This is a system of two equations with three unknowns, and it has an infinitude of solutions, and there is no constant value for $a+b+c$. However, $a+b+2c=8$. If this was a textbook exercise, are you sure it was presented asking for $a+b+c$, not $a+b+2c$?
