We cannot multiply directly to remove the 22 from the denominator of the argument of sine. Instead, we will let theta = x/2θ=x2, then solve for thetaθ, and substitute back in after we have gotten rid of the sine function.
Let theta = x/2θ=x2
=> 2sqrt(3)sin(theta) = 3⇒2√3sin(θ)=3
=> sin(theta) = 3/(2sqrt(3)) = sqrt(3)/2⇒sin(θ)=32√3=√32
Checking our unit circle, we find that sin(theta) = sqrt(3)/2sin(θ)=√32 when theta = pi/3θ=π3 or theta = (2pi)/3θ=2π3. As sine is periodic with a period of 2pi2π, we can add any integer multiple nn of 2pi2π to one of these without changing the value of sinesine. So we have:
theta = pi/3 + 2npiθ=π3+2nπ
or
theta = 2(pi)/3 + 2npiθ=2π3+2nπ
Now that the sine is removed, let's substitute x/2x2 back in.
x/2 = pi/3 + 2npix2=π3+2nπ
or
x/2 = (2pi)/3 + 2npix2=2π3+2nπ
Multiplying by 22 we get our desired result:
x = (2pi)/3 + 4npix=2π3+4nπ
or
x = (4pi)/3 + 4npix=4π3+4nπ
:. x in {(2pi)/3 + 4npi, (4pi)/3 + 4npi}
