I'm getting **2** answers:

\(\displaystyle sin\,\theta= \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\)

I started by using the tangent double identity:

\(\displaystyle \frac{2tan\,\theta}{1-tan^2\,\theta}=\frac{4}{3}\)

Cross-multiplied:

\(\displaystyle 3\left(2tan\,\theta \right)=4\left(1-tan^2\,\theta \right)\)

... solved for \(\displaystyle tan\,\theta\), then rewrote in terms of \(\displaystyle sin\,\theta\).

EDIT: I originally thought that both answers I wrote above were plus-minus, but I forgot that \(\displaystyle tan\,2\theta\) is positive, which means that the angle \(\displaystyle 2\theta\) has to be in Quadrants I or III. So \(\displaystyle \theta\) is between 0 and 45° or between 90° and 135°, and for those values of \(\displaystyle \theta\), sine is positive.