If $\sin x+\sin^{2} x=1$ , Find $\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $ 
If $\sin x+\sin^{2} x=1$, then the value of 
  $\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
  is equal to 
$a.)\ 0 \\
b.)\ 1 \\
c.)\ 2 \\
\color{green}{d.)\ \sin^{2} x} $

$\boxed{\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2\\
=\cos^{6} x\left(\cos^{6} x+3\cos^{4} x+3\cos^{2} x+1\right)+2\sin^{2} x+\sin x-2\\
=\sin^{3} x\left(\cos^{2} x+1\right)^{3}+2\sin^{2} x+\sin x-2\\
=\sin^{3} x\left(\sin x+1\right)^{3}+2\sin^{2} x+\sin x-2\\
=\left(\sin x[\sin x+1]\right)^{3}+2\sin^{2} x+\sin x-2\\
=\left(\sin^{2} x+\sin x\right)^{3}+2\sin^{2} x+\sin x-2\\
=1+\sin^{2} x+\sin^{2} x+\sin x-2\\
=\sin^{2} x} $
I found this solution but considering the time as $1-2$ min to solve this question,
I am looking for a short and simple way.
I have studied maths up to $12$th grade

Observe that $\sin x = 1-\sin^2 x =\cos^2 x$. Moreover let $\cos^4 x =t$ then the expression given is
\begin{align*}
 & = \color{red}{t^3+3t^2\sqrt{t}+3t(\sqrt{t})^2+(\sqrt{t})^3}+2t+\sqrt{t}-2\\
&=\color{red}{(t+\sqrt{t})^3}+2t+\sqrt{t}-2\\
&=(\cos^4x+\cos^2x)^3+2\cos^4x+\cos^2x-2\\
\text{using the fact that $\sin x =\cos^2 x$}\\
&=(\sin^2x+\cos^2x)^3+2\sin^2x+\cos^2x-2\\
&=1+1+\sin^2x-2\\
&=\sin^2x.
\end{align*}
