If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$ If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$
My Attempt:
$$\sin x + \sin^2 x=1$$
$$\sin x = 1-\sin^2 x$$
$$\sin x = \cos^2 x$$
Now, 
$$\cos^8 x + 2\cos^6 x + \cos^4 x$$
$$=\sin^4 x + 2\sin^3 x +\sin^2 x$$
$$=\sin^4 x + \sin^3 x + \sin^3 x + \sin^2 x$$
$$=\sin^3 x(\sin x +1) +\sin^2 x(\sin x +1)$$
$$=(\sin x +1) (\sin^3 x +\sin^2 x)$$
How do I proceed further?

\begin{align}
& \cos^8 x + 2\cos^6 x + \cos^4 x \\[8pt]
= {} & \sin^4 x + 2\sin^3 x + \sin^2 x \\[8pt]
= {} & (\sin^2 x)^2 + 2(\sin^2 x)(\sin x) + \sin^2 x \\[8pt]
= {} & (1-\cos^2 x)^2 + 2(1-\cos^2 x)(\sin x) + (1-\cos^2 x) \\[8pt]
= {} & (1-\sin x)^2 + 2(1 - \sin x)(\sin x) + (1-\sin x) \\[8pt]
= {} & (1-2\sin x + \sin^2 x) + 2(\sin x - \sin^2 x) + (1-\sin x) \\[8pt]
= {} & 2 - \sin x -\sin^2 x \\[8pt]
= {} & 2 - \sin x -(1-\cos^2 x) \\[8pt]
= {} & 2 - \sin x - (1-\sin x) \\[8pt]
= {} & 1.
\end{align}
