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?

Here is a shorter way to do it. Your condition can be rewritten in the forms
$$\sin x = \cos^2(x) \implies \sin^n(x) = \cos^{2n}(x) \implies \sin^n(x)+\sin^{n-1}(x) = \sin^{n-2}(x)$$
We thus rewrite your cosines as
$$\begin{align}
\cos^8 x + 2\cos^6 x + \cos^4 x &= \sin^4(x)+2\sin^3(x)+\sin^2(x)\\ &= [\sin^4(x)+\sin^3(x)]+[\sin^3(x)+\sin^2(x)]\\ &= \sin^2(x)+\sin(x)\\&=1
\end{align}
$$
