The correct option is B 2
(81)sin2x+(81)1−sin2x=30(81)sin2x+81(81)sin2x=30Let (81)sin2x=tt+81t=30⇒t2+81=30t⇒t2−30t+81=0⇒t2−27t−3t+81=0⇒(t−3)(t−27)=0⇒t=3,27⇒(81)sin2x=3,33⇒34sin2x=31,33⇒4sin2x=1,3⇒sin2x=14,34in [0,π], sinx≥0sinx=12,√32x=π6,5π6,π3,2π3
Number of solutions =4