A generator at a regional Power Station produces an alternating voltage, according to the function, 𝑉 below:

i) We would like to solve the equation 
\[240 \sin (20t)=120  \Rightarrow  \sin (2t)=\frac{120}{240}=\frac{1}{2}.\]
Hence
\[\sin (2t)=\frac{1}{2}   \Rightarrow    2t=2k \pi\pm \frac{\pi}{6}   \Rightarrow  t=k \pi \pm \frac{\pi}{12},  k=0, \pm 1, \pm 2, \dots .\]
So the solutions are
\[t=\pm \frac{\pi}{12}, \pi \pm \frac{\pi}{12}, 2\pi \pm \frac{\pi}{12}, 3 \pi \pm \frac{\pi}{12}, \dots .\]

We only want the solutions that lie in $[0,2\pi]$ which are
\[\frac{\pi}{12}, \pi \pm \frac{\pi}{12}, 2\pi -\frac{\pi}{12}\]
as the remaining solutions are outside $[0, 2\pi]$. So there are a total of 4 solu
tions: 
\[\frac{\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{23\pi}{12}.\]

(ii) The exact times are indeed 

\[\frac{\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{23\pi}{12}.\]