solve sin(2x)=0.5. between 0<x<2pi

1)Take the inverse sin to take x from the sin(2x):

2x=arcsin(0.5).

2)Evaluate arcsin(0.5) to get pi/6:

so 2x= pi/6

3)Dividing by 2 to simplify we get 

x=pi/12.

4)To find the second solution we note that (pi/2)-(pi/12) =(5pi/12) is also a solution. 

So x= (5pi/12)

5)Sin(2x) has a period of pi. So to find the rest of the solutions we add pi to our previous solutions. 

So now x=pi/12, 5pi/12, 13pi/12 , 17pi/12