Question

How do I solve for x in degrees given sin 2x - 1 = 0?

Answer 1

`x=45°+180°n` where n is any integer.

Explanation:

Replace `2x` with `u`

`sin(u)-1=0`
`sin(u)=1`

`u=pi/2+2npi` where n is any integer.

We know that `pi` gives us `1` for `sin(u)`; however, since we're not told that we're working on an interval, we must account for all other possible solutions (there are infinitely many values that give `1` that repeat on a period of `2pi` , such as `+-(5pi)/2, +-(9pi)/2, +-(13pi)/2`) by adding an integer multiplied by `2pi` to our solution.

Switch back to `2x`:

`2x=pi/2+2npi`

Simplify so that we only have an `x` on the lefthand side :
`x=(pi/2+2npi)/2`

`x=pi/4+npi`

To convert radians by degrees, multiply by `180/pi`

`cancel(pi)/4 * 180/cancel(pi) = 180/4=45°`

`(ncancel(pi))*180/cancel(pi)=(180n)` where n is any integer.

Add these two back together:

`x=45°+180°n`

Answer 2

I am not sure whether you mean `sin(2x - 1)` or `sin (2x) - 1`, so I'm going to do both.

1) `sin(2x - 1) = 0`

However, we know that `sin(npi) = 0`, for any integer `n`.

This implies `2x-1 = npi`, therefore

`x = (npi+1)/2`, any `n -> "Z"`

and in degrees :

`x = (180°n + 1)/2`, any `n -> "Z"`.

2) `sin(2x) - 1 = 0`

`sin(2x) = 1`

We know that `sin(((2n-1)pi)/2) = 1`, again, for any integer `n`.

This implies `2x = ((2n-1)pi)/2`, therefore

`x = ((2n-1)pi)/4`, any `n -> "Z"`.

and in degrees :

`x = ((2n-1)180°)/4`, any `n-> "Z"`.