solve for x between 0 < X < 2π
Show work please
Show work please

sin x ( sin x  1 ) = 0
sin x = 0 , sin x = 1
x = 0 , π , 2π , π/2 for 0 ≤ x ≤ 2π
sin x = 0 , sin x = 1
x = 0 , π , 2π , π/2 for 0 ≤ x ≤ 2π

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sin^2(x) = sin(x)
=> sin^2(x)  sin(x) = 0
=> sin(x)[sin(x)  1] = 0
case 1: sin(x) = 0
=> x = Π
case 2: sin(x)  1 = 0
=> sin(x) = 1
=> x = Π/2
x = Π/2 and Π
=> sin^2(x)  sin(x) = 0
=> sin(x)[sin(x)  1] = 0
case 1: sin(x) = 0
=> x = Π
case 2: sin(x)  1 = 0
=> sin(x) = 1
=> x = Π/2
x = Π/2 and Π