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}#
