<s>[INST] 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
 [/INST] Hint: $\sin x = 1 - \sin^2 x = \cos^2x \Rightarrow \sin^2 x = \cos^4 x \Rightarrow \cos^4 x = 1 - \cos^2 x$. Thus you can express $\cos^{10}x , \cos^8 x, \cos^6 x, \cos^4x$ in term of $\cos^2 x$.
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