According to the double angle formula sin2α=2sinαcosα, what is the solution of the domain of this problem? Sin2θ > 0, and then we can get 2θ∈(2kπ, (2k+ 1)π) from the sine function value domain, then θ∈(kπ, (k+? ) π), k values are different (0,? π) and (π, 3π/2) are both reasonable domains, but you have miscalculated yourself.