find exactly all the solutions in [0, 2pi) to 4 sin(x)cos(x) = 1

Answer 1

you have 2sin2x = 1
so, sin2x = 1/2
2x = pi/6 + 2n*pi
x = pi/12 + n*pi = pi/12, 13pi/12

Answer 2

from sin 2x = 1/2
2x = π/6 or 2x = π-π/6 = 5π/6

x = π/12 or 5π/12
but the period of sin 2x is π
so we get 2 more solutions:
x = π/12 + π = 13π/12
or
x = 5π/12 + π = 17π/12

in [ 0 , 2π]
x = π/12 , 5π/12 , 13π/12 and 17π/12

Answer 3

good work; I missed out on that one!