So $2^{\circ}$ part will be covered in a time $t=\frac{T}{2}=1$ sec

For the half $1^{\circ}$ part

$\theta=\theta_{2} \sin (\omega t)$

$1^{\circ}=2^{\circ} \sin \left(\frac{2 \pi}{T} \times t\right) \Rightarrow \frac{t}{2}=\sin \left(\frac{2 \pi}{T} \times t\right)$

$\Rightarrow \frac{\pi}{6}=\pi \times 1 \Rightarrow t=1 / 6 \mathrm{sec}$

Total time $\Rightarrow \frac{T}{2}+2 t$

$\Rightarrow 1+2 \times \frac{1}{6}=1+\frac{1}{3}=\frac{4}{3} \mathrm{sec}$