We have,
$$
\begin{array}{ll}
& \sin x+\sin ^{2} x=1 \\
\Rightarrow \quad & \sin x=1-\sin ^{2} x \\
\Rightarrow \quad & \sin x=\cos ^{2} x \\
\text { Now, } \cos ^{12} x & +3 \cos ^{10} x+3 \cos ^{8} x+\cos ^{6} x \\
& =\sin ^{6} x+3 \sin ^{5} x+3 \sin ^{4} x+\sin ^{3} x \\
& =\sin ^{3} x\left[\sin ^{3} x+3 \sin ^{2} x+3 \sin x+1\right] \\
& =\sin ^{3} x(\sin x+1)^{3} \\
& =\left(\sin ^{2} x+\sin x\right)^{3} \\
& =(1)^{3}=1
\end{array}
$$