If sin⁡x+sin2⁡x=1 then the value of cos12⁡x+3cos10⁡x+3cos8⁡x+cos6⁡x−2 is equal to

We have  sin⁡x+sin2⁡x=1or  sin⁡x=1−sin2⁡x or sin⁡x=cos2⁡xnow  cos12⁡x+3cos10⁡x+3cos8⁡x+cos6⁡x−2=sin6⁡x+3sin5⁡x+3sin4⁡x+sin3⁡x−2=sin2⁡x3+3sin2⁡x2sin⁡x+3sin2⁡x(sin⁡x)2+(sin⁡x)3−2=sin2⁡x+sin⁡x3−2=(1)3−2=−1