Question

`sin(x) > 1/2" in "[0,2pi)`?

How do I solve for `x`

Answer 1

`pi/6 < x < {5pi}/6`

Explanation:

Keep in mind that `sinx` is always greater than or equal to zero and less than or equal to 1.
`0 leq sinx leq 1`. That being said we have the following range: `1 geq sinx > 1/2`

Looking at the unit circle we can see that when `sinx =1/2`, `x=pi/6` or `x={5pi}/6`. This gives us a range of `pi/6 < x < {5pi}/6`

When `sinx=1`, `x = pi/2`
This is already within the range of `pi/6 < x < {5pi}/6` so we do not need to explicitly include that in our range.