Solve the equation on the interval [0, 2pi) 2sin2 x = 4sin x + 6 Solution sin x = t 2t^2 - 4t - 6 = 0 => t^2 -2t -3 = 0 => t^2 -3t + t -3 = 0 => t(t-3) +1(t-3) = 0 => (t-3)(t+1) = 0 => t = 3, -1 But sin t cannot be 3 so , sinx = -1 in this given interval x = 3*PI / 2.