Multiply numerator and denominator by the conjugate of the denominator $(1+2i\sin\theta)$:

Numerator real part: $(2+3i\sin\theta)(1+2i\sin\theta)$ real part $= 2-6\sin^2\theta$

For purely imaginary, the real part must be zero: $2-6\sin^2\theta=0\Rightarrow\sin^2\theta=\frac{1}{3}\Rightarrow\sin\theta=\frac{1}{\sqrt{3}}$

Therefore $\theta=\sin^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right)$