दिया है , $\sin x + {\sin ^2}x = 1$

या  $\sin x = 1 – {\sin ^2}x$ या $\sin x = {\cos ^2}x$

$\therefore$ ${\cos ^{12}}x + 3{\cos ^{10}}x + 3{\cos ^8}x + {\cos ^6}x – 2$

$ = {\sin ^6}x + 3{\sin ^5}x + 3{\sin ^4}x + {\sin ^3}x – 2$

$ = {({\sin ^2}x)^3} + 3{({\sin ^2}x)^2}\sin x  + 3({\sin ^2}x){(\sin x)^2} + {(\sin x)^3} – 2$

$ = {({\sin ^2}x + \sin x)^3} – 2$

$ = {(1)^3} – 2$                     $[ \because \sin x + {\sin ^2}x = 1$ (दिया है)$]$ 

$= -1.$