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:
Solution
$\sin x = \cos^2 x$(Putting in equation)
$\sin^6x + 3 \sin^5 x + 3\sin^4 x + \sin^4 x + 2 \sin^2 x + \sin x - 2$
$= \sin^4 x (\sin^ x + \sin x) + 2 \sin 3x(\sin^2 x + \sin x) + \sin^2 x (\sin^2 x + \sin x) - \sin x 2$
$= \sin^2 x(\sin^2 x + \sin x) + \sin x(\sin^2 x + \sin x) - \sin x$
$= \sin^2 x + \sin x - \sin x - \sin^2 x$