Sines amplifies cosines

We have $\sin x = 1- \sin^2 x = \cos^2 x$

$\begin{aligned} & & \cos^{12} x + 3 \cos^{10} x + 3 \cos^8 x + \cos^6 x - 2 \\ & = & \left ( \cos^2 x \right )^6 + 3 \left ( \cos^2 x \right )^5 + 3\left ( \cos^2 x \right )^4 + \left ( \cos^2 x \right )^3 - 2 \\ & = & \sin^6 x + 3\sin^5 x + 3\sin^4 x + \sin^3 x - 2 \\ & = & \sin^3 x \left (\sin^3 x + 3\sin^2 x+ 3\sin x + 1 \right ) - 2 \\ & = & \sin^3 x \ \left (\sin x + 1 \right )^3 - 2 \\ & = & \left ( \sin x (\sin x + 1 ) \right )^3 - 2 \\ & = & \left ( \sin^2 x + \sin x \right )^3 - 2 \\ & = & 1^3 - 2 = \boxed{-1} \\ \end{aligned}$