Find an angle θ, where 0 < θ < (π)/2, which increases twice as fast a…

The correct answer is Option (3) → $\frac{\pi}{3}$ ##

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

$\Rightarrow \frac{d\theta}{dt} = 2 \frac{d}{dt}(\sin \theta)$

Now, on differentiating both sides w.r.t. $t$, we get

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

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

$∴\theta = \frac{\pi}{3}$

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