Find an angle $\theta$, where $0 < \theta < \frac{\pi }{2}$, which increases twice as fast as its sine.

Let $\theta$ increases twice as fast as its sine.

$\Rightarrow$ $\theta = 2\sin \theta$

By differentiating both sides w.r.t. $t$, we get

$\frac{{d\theta }}{{dt}} = 2 \cdot \cos \theta \cdot \frac{{d\theta }}{{dt}} \Rightarrow 1 = 2\cos \theta$

$\Rightarrow$ $\frac{1}{2} = \cos \theta \Rightarrow \cos \theta = \cos \frac{\pi }{3}$

Therefore,$\theta = \frac{\pi }{3}$

So, the required angle is $\frac{\pi }{3}$.