We start by adding 1 to both sides:
2sin(2x)-cancel(1+1)=0+1
2sin(2x)=1
Next we divide both sides by 2:
(cancel2sin(2x))/cancel2=1/2
sin(2x)=1/2
We know that sin(pi/6+2pik)=1/2, which gives:
2x=pi/6+2pik
(cancel2x)/cancel2=((pi/6+2pik)/2)
x=pi/12+pik
This is one of the solutions. We can find the other by using the identity sin(theta)=sin(pi-theta). This means that sin(pi-pi/6+2pik) also equals 1/2, which gives the solution:
2x=pi-pi/6+2pik
2x=(5pi)/6+2pik
(cancel2x)/cancel2=(5pi)/(2*6)+(2pik)/2
x=(5pi)/12+pik
Now we need to find which of these solutions are inside the desired interval. We can write them out, seeing which ones are smaller than 2pi:
pi/12+pik->pi/12, (13pi)/12,(25pi)/12...
(5pi)/12+pik->(5pi)/12, (17pi)/12, (29pi)/12
Everything larger than (24pi)/12=2pi, is not in the interval we want, so we can ignore those, leaving us with the following solutions:
x=pi/12, x=(5pi)/12, x=(13pi)/12, x=(17pi)/12