Given, 3sin2x+cos2x+1+32−sin2x−cos2x=28
Taking 3sin2x+cos2x+1=y,
y+27y=28 or y2−28y+27=0 or y=1,27=3o,33
∴sin2x+cos2x+1=0,3
But the maximum value of sin2x+cos2x=√2.
∴ we get only sin2x+cos2x+1=0
⇒2tanx1+tan2x+1−tan2x1+tan2x+1=0⇒tanx=−1
Also, sin2x+cos2x+1=0
⇒2sinxcosx+cos2x=0⇒cosx=0 or 2sinx+cosx=0
cosx=0 satisfies the equation, but tanx=−12 does not exist.