Sum of the roots of 3sin2x+2cos2x+31−sin2x+2sin2x=28 in [−2π,2π] is
π
3sin2x+2cos2x+31−sin2x+2sin2x=283sin2x+2cos2x+31−sin2x+2(1−cos2x)=283sin2x+2cos2x+33.3−sin2x−2cos2x=28
Let t=3sin2x+2cos2x⇒t+33t=28⇒t2−28t+27=0t=1, 27sin2x+2cos2x=0 or sin2x+2cos2x=32 sinx cosx+2cos2x=0 i.e sin2x=1, 2cos2x=1 [which is not possible. ]
2cosx(sinx+cosx)=0cosx=0, sinx+cosx=0
cosx=0 ⇒x=π2,−π2,3π2,−3π2.sinx+cosx=0⇒tanx=−1⇒x=−π4,3π4,7π4,−5π4∴sum=π