The sum of distinct solutions of the equation 4tanx+tan2x+1=22sinx+π41+tan2x in the interval 0,π is:
π2
4π3
2π
7π6
4tanx+sec2x=22sinx+π4sec2x⇒4sinxcosx+1=2sinx+cosx
⇒ 2sinx2cosx-1-2cosx-1=0
sinx=12,cosx =12⇒x=π6,5π6,π3