$Solution:$ Given, $\sin x=1-\sin ^2 x$
$$\begin{array}{l}\Rightarrow \quad \sin x=\cos ^2 x \\\text { Now, } \cos ^8 x+2 \cos ^6 x+\cos ^4 x \\=\left(\cos ^2 x\right)^4+2\left(\cos ^2 x\right)^3+\left(\cos ^2 x\right)^2 \\=\sin ^4 x+2 \sin ^3 x+\sin ^2 x \quad \text { [from Eq. (i)] } \\=\sin ^2 x[\sin x+1]^2 \\=(1-\sin x)(1+\sin x)(1+\sin x) \\\end{array}$$
$$\begin{array}{l}=\cos ^2 x(1+\sin x)=\sin x(1+\sin x) \\=\sin x+\sin ^2 x \\=1 \quad \text { [from } \mathrm{Eq} .(\mathrm{i})]\end{array}$$
$Correct$ $Option$ $is$ ($d$).
