Find all solutions to the equation sin2x+sinx-2cosx-1=0 in the interval [0, 2pi)

The solutions appear to be {π/2, 2π/3, 4π/3}.

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Replacing sin(2x) with 2sin(x)cos(x), you have
  2sin(x)cos(x) +sin(x) -2cos(x) -1 = 0
  sin(x)(2cos(x) +1) -(2cos(x) +1) = 0 . . . . factor by grouping
  (sin(x) -1)(2cos(x) +1) = 0

This has solutions
  sin(x) = 1
  x = π/2
and
  2cos(x) = -1
  cos(x) = -1/2
  x = {2π/3, 4π/3}