When `sinx=1/2`, the two values of x between 0 and `2pi` and `pi//6` and `5pi//6`
From the graph of `y=sinx`, it is obvious that between 0 and `2pi`,
`sinx gt1/2` for `pi/6 lt x lt 5pi//6`
Here, `sin x gt 1//2`
`rArr 2npi + pi//6 lt x lt 2npi +5pi//6, n in I`
Thus, the required solution set is `underset(n in I)cup(2npi+pi/6, 2npi+(5pi)/6)` Ans.
