C-08-Q164mediumsingle_mcqIn 0<θ<π20 < \theta < \dfrac{\pi}{2}0<θ<2π, the equation 6sin2θ−11sinθ+4=06\sin^{2}\theta - 11\sin\theta + 4 = 06sin2θ−11sinθ+4=0 yields sinθ\sin\thetasinθ equal to —a12\dfrac{1}{2}21b43\dfrac{4}{3}