Find all solutions in the interval [0, 2π).2 sin2x = sin x x = pi divided by three. , two pi divided by three. x = pi divided by two. , three pi divided by two. , pi divided by three. , two pi divided by three. x = 0, π, pi divided by six , five pi divided by six x = pi divided by six , five pi divided by six

bro, use your trig identities 
[tex]sin2x=2sinxcosx[/tex]
So,
[tex]2sin2x=sinx[/tex]
[tex]4sinxcosx=sinx[/tex]
[tex]4sinxcosx-sinx=0[/tex]
[tex]sinx(4cosx-1)=0[/tex]
[tex]sinx=0[/tex] & [tex] 4cosx=1[/tex]

x = 0, π

x = acos(1/4)=1.32