We cannot multiply directly to remove the 2 from the denominator of the argument of sine. Instead, we will let theta = x/2, then solve for theta, and substitute back in after we have gotten rid of the sine function.
Let theta = x/2
=> 2sqrt(3)sin(theta) = 3
=> sin(theta) = 3/(2sqrt(3)) = sqrt(3)/2
Checking our unit circle, we find that sin(theta) = sqrt(3)/2 when theta = pi/3 or theta = (2pi)/3. As sine is periodic with a period of 2pi, we can add any integer multiple n of 2pi to one of these without changing the value of sine. So we have:
theta = pi/3 + 2npi
or
theta = 2(pi)/3 + 2npi
Now that the sine is removed, let's substitute x/2 back in.
x/2 = pi/3 + 2npi
or
x/2 = (2pi)/3 + 2npi
Multiplying by 2 we get our desired result:
x = (2pi)/3 + 4npi
or
x = (4pi)/3 + 4npi
:. x in {(2pi)/3 + 4npi, (4pi)/3 + 4npi}
