If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$ 
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$

My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\big(1+\sin^2x\big)\\
&=\sin x\cdot(1+\sin^2x)\bigg[\sin^2x\cdot(1+\sin^2x)^2-4\sin x\cdot(1+\sin^2x)+8\bigg]\\
&=
\end{align}
I don't think its getting anywhere with my attempt, so how do I solve it ?
Or is it possible to get the $x$ value that satisfies the given condition $\sin x+\sin^2x+\sin^3x=1$ ?
Note: The solution given in my reference is $4$.
 A: Let us denote $y = \sin x$. The relation we have for $y$ is then $y + y^2 + y^3 = 1$, or also if we multiply by $y-1$, we get $y^4 = 2y - 1$. The idea is simply to write the expression in $\cos^2 x$ given in terms of $y$, and use the relation to simplify it. We have
\begin{align*}
\cos^6x-4\cos^4x+8\cos^2x 
  &= (1 - y^2)^3 - 4 (1 - y^2)^2 + 8 (1 - y^2) \\
  &= (1 - y^2) [(1 - 2y^2 + y^4) + (4y^2 - 4) + 8] \\
  &= (1 - y^2) [5 + 2y^2 + y^4] \\
  &= (1 - y^2) [5 + 2y^2 + (2y - 1)] \\
  &= 2(1 - y^2) [2 + y + y^2] \\
  &= 2 [2 + y + y^2 - 2y^2 - y^3 - y^4] \\
  &= 2 [2 + y + (-y)(y + y^2 + y^3)] \\
  &= 2 [2 + y - y] \\
  &= 4.
\end{align*}
Maybe there is some approach that is more straightforward using clever algebraic manipulation. However, this solution is quite clear from a theoretical point of view: you have a polinomial in $y$ that you want to simplify using the relation given. Then, you can divide it by the polynomial given in the relation and the remainder will be a polynomial with degree at most 2. In this specific case, it was the constant polynomial 4.
A: $$\sin(x)(1+\sin^2(x))=1-\sin^2(x)$$
Square both sides & replace $\sin^2(x)$ with $1-\cos^2(x)$
A: Given $$\sin x+\sin^2x+\sin^3x=1$$$$\sin x+\sin^3x=1-\sin^2x$$$$(\sin x+\sin^3x)^2=(1-\sin^2x)^2$$
$$\sin^2x+\sin^6x+2\sin^4x=\cos^4x$$
$$1-\cos^2x+(1-\cos^2x)^3+2(1-\cos^2x)^2=\cos^4x$$
$$1-\cos^2x+1-3\cos^2x+3\cos^4x-\cos^6x+1-4\cos^2x+2\cos^4x=\cos^4x$$$$\cos^6x-4\cos^4x+8\cos^2x=4$$
A: Write $s := \sin x, c := \cos x$.
Hint Using $s^2 + c^2 = 1$, we can write our expression as
$$c^6 - 4 c^4 + 8 c^2 = (1 - s^2)^3 - 4 (1 - s^2)^2 + 8 (1 - s^2) = -s^6 - s^4 - 3 s^2 + 5 .$$
Now, perform polynomial long division by $s^3 + s^2 + s - 1$.

 Carrying out long division gives that our expression is $$(s^3 + s^2 + s - 1) p(s) + 4$$ for some cubic polynomial $p$. But we're given that $s^3 + s^2 + s = 1$, so the first term is zero, and thus $$\color{#df0000}{\boxed{c^6 - 4 c^4 + 8 c^2 = 4}} .$$

A: Let $t=\sin x$ and solve the cubic $$t^3+t^2+t=1$$
Wolfram Alpha gives the real solution as 
$$t=(1/3) (-1 - 2/(17 + 3 \sqrt {33})^{1/3} + (17 + 3 \sqrt{33})^{1/3})$$
Plug the real solution of the above to get  $$(1-t^2)^3 -4(1-t^2)^2+8(1-t^2) =4$$ 
A: Start off with the identity $\sin^2(x)+\cos^2(x)=1$. Rearrange this to find $\sin(x)=\pm\sqrt{1-\cos^2(x)}$. Substitute this into your equation and then rearrange to find $\sqrt{1-\cos^2(x)}$. You should be able to take it from there.
Note that it's not an identity that $\cos^2(x)=\sin(x)+\sin^3(x)$.
