How to find the solution for this trig equation (pre-calc)?

Question

2sin^2(x) + sin(x) - 1 = 0 (each answer in the form θ + 2πk, 0 ≤ θ < 2π)
I found two solutions which are: π/2 + 2πk , 5π/2 + 2πk ... there seems to be one more solution but I can't figure it out...

Answer 1

Given that 2sin^2(x) + sin(x) - 1 = 0

Let us assume sin(x) = t

          => 2t^2 + t - 1 = 0                       [ Here a = 2, b = 1 and c = -1 ]

          => t = (-1 +/- sqrt(1^2 - 4*2*-1)) / 4      [ By quadratic formula -b +/- sqrt(b^2 - 4ac) / 2a ]

          => t = (-1 +/- sqrt(1+8)) / 4

          => t = (-1 +/- sqrt(9)) / 4     

          => t = (-1 +/- 3) / 4                     [ Since sqrt(9) = 3 ]

          => t = -4/4 , 2/4

          => t = -1 , 1/2

          => sin(x) = -1 , 1/2                     [ Since t = sinx ]

          => x = 3π/2 , π/6 , 5π/6

But we have to write in the form θ + 2πk

          => x = 3π/2 + 2π * k , π/6 + 2π * k , 5π/6 + 2π * k