Given that: $\sin x+\sin^2x=1\cdots\cdots\to(i)$
Above equation can be written as:
$\implies \sin x=1-\sin^2x$
$\because \sin^2x+\cos^2x=1$
$\implies \sin x=\cos^2x \cdots\cdots\cdots \to(ii)$
Now we have to find the value of $\cos^8x+2\cos^6x+\cos^4x$
$\implies \cos^8x+2\cos^6x+\cos^4x$
$\implies (\cos^4x)^2+2*\cos^4x*\cos^2x+(cos^2x)^2$
$\because \left [ (a+b)^2=a^2+b^2+2ab\right ]$
$\implies (\cos^4x+\cos^2x)^2$
From eq. $(i)$, $\sin x=\cos^2 x, \sin^2x = \cos^4x$
$\implies (\sin^2x+\sin x)^2$
$\implies 1^2$ $ \cdots\cdots \to(1)$
$\implies1$
Option $(C)$ is correct.