Test the identity:
Let θ=30°, then sinθ=½ and cosθ=√3/2, so 2sin2θ=2/4=½; and 3(1-cosθ)=(3/2)(2-√3).
So the identity is not true. Let's see what is true:
2sin2θ=2(1-cos2θ)=2(1-cosθ)(1+cosθ).
3(1-cosθ)=3(1-(1-2sin2(½θ))=6sin2(½θ), so:
6sin2(½θ)=3(1-cosθ), sin2(½θ)=½(1-cosθ).
If 2sin2θ=3(1-cosθ) is an equation instead of an identity it can be solved for θ.
2sin2θ=2(1-cos2θ)=3(1-cosθ),
2-2cos2θ=3-3cosθ,
2cos2θ-3cosθ+1=(2cosθ-1)(cosθ-1), so cosθ=½ or 1.
If θ is between 0° and 360°, the solution is θ=0°, 60° or 300°.