So I have the problem
2sin(3x)-1 = 0 find all solutions in the interval [0,2π)

I have gotten
3x =
π/6 + 2kpi
and
5π/6 + 2kpi
and
x =
π/18 + 2kpi/3
and
5π/18 +2kpi/3

I have taken the inequality of π/18 + 2kπ/3

and have gotten
-1/12=

Question

So I have the problem 
2sin(3x)-1 = 0 find all solutions in the interval [0,2π)

I have gotten 
3x = 
π/6 + 2kpi 
and
5π/6 + 2kpi
and
x = 
π/18 + 2kpi/3
and
5π/18 +2kpi/3

I have taken the inequality of π/18 + 2kπ/3

and have gotten 
-1/12=<K<35/12

-----The Part I do not understand----

Now it states based on the solution to that inequality that

k=0, k=1 and k=2 
and thus the solutions are
x= π/18 , 13π/18 ,25π/18

NOTE: π = pi

anonymous · Answer

Can someone explain to me how the k values and possible solutions were obtained

anonymous · Answer

Also 
k = an element of any integer

anonymous · Answer

its probably because for k greater than or equal to 3, you obtain numbers than are bigger than 2pi.  the question is only concerned about the solutions from [0,2pi)

anonymous · Answer

So I would computer 35/12 = 2.9166
and just use k values less than that number but how do I obtain the values

13π/18 ,25π/18

anonymous · Answer

Ok I got it I just sub the k value into the solution π/18 + 2kpi/3 and solve :D 
Thanks for helping me see the solution