Find all solutions of 2sin(x)cos(x) + 2sin(x) - cos(x) - 1 = 0

where 0 <= x < 2pi answers in radians also, can this be done without substitution, just factorising and such? Thanks I'm stuck on this TT

You can factor by grouping:
2sin(x) (cos(x) + 1) - 1(cos(x) + 1) = 0
(2sin(x) - 1)(cos(x) + 1) = 0
2sin(x) - 1 =0
2sin(x) = 1
sin(x) = 1/2
x = pi/6, 5pi/6
also:
cos(x) + 1 = 0
cos(x) = -1
x = pi