Question

What are the all the solutions between 0 and 2π for `sin2x-1=0`?

Answer 1

`x = pi/4` or `x = (5pi)/4`

Explanation:

`sin(2x) - 1 = 0`

`=> sin(2x) = 1`

`sin(theta) = 1` if and only if `theta = pi/2+2npi` for `n in ZZ`

`=>2x = pi/2+2npi`

`=> x = pi/4+npi`

Restricted to `[0, 2pi)` we have `n=0` or `n=1`, giving us

`x = pi/4` or `x = (5pi)/4`

Answer 2

`S={pi/4,5pi/4}`

Explanation:

First, isolate the sine

`sin(2x)=1`

Now, take a look at your unit circle


Now, the sine correspond to the `y` axis, so we can see that the only point between `0` and `2pi` where the sine is `1` is `pi/2` radians, so we have:

`2x = pi/2`

We want to solve for x, so

`x = pi/4`

However, remember that the period of the normal sine wave is `2pi`, but since we're working with `sin(2x)`, the period has changed; basically what we know is that there is a constant `k` that will act as the period, so:

`2(pi/4 + k) = pi/2 + 2pi`
`pi/2 + 2k = pi/2 + 2pi`
`2k = 2pi`
`k = pi`

And since `pi/4 + pi` or `5pi/4` is between `0` and `2pi`, that enters our set of solutions.

`S={pi/4,5pi/4}`