Question

How do you solve `sin x = 1/2` for x in the interval [0,2pi)?

Answer 1

Hence

`sinx=1/2=>x_1=pi (2 n+1/6) or x_2=pi (2 n+5/6)`

where `n` is an integer

But hence `x` belongs to `[0,2pi)` the acceptable solutions are

`x_1=pi/6` and `x_2=5*pi/6`