If 2cosx+2cos3x=cosy,2sinx+2sin3x=siny then the value of cos2x of
– 7/8
1/8
– 1/8
7/8
cosy=2×2cos 2xcos x
siny=2×2sin2xcosx
⇒1=cos2y+sin2y=16cos2x
So cos2x=2cos2x−1=−7/8