If sin x + sin

We have:

$\sin x+\sin ^{2} x=1$

$\Rightarrow \sin x=1-\sin ^{2} x$

$\Rightarrow \sin x=\cos ^{2} x$           ...(2)

Now, taking square of $(1):$

$\Rightarrow\left(\sin x+\sin ^{2} x\right)^{2}=(1)^{2}$

$\Rightarrow(\sin x)^{2}+\left(\sin ^{2} x\right)^{2}+2(\sin x)\left(\sin ^{2} x\right)=1$

$\Rightarrow(\sin x)^{2}+(\sin x)^{4}+2(\sin x)^{3}=1$

$\Rightarrow(\sin x)^{2}+2(\sin x)^{3}+(\sin x)^{4}=1$

$\Rightarrow\left(\cos ^{2} x\right)^{2}+2\left(\cos ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{4}=1$

$\Rightarrow \cos ^{4} x+2 \cos ^{6} x+\cos ^{8} x=1$

$\therefore \cos ^{8} x+2 \cos ^{6} x+\cos ^{4} x=1$