Question

How do I solve the following math equation?

`1=sqrt2sin(pi/2x+(5pi)/4)` for [0,6)

Answer 1

`x=2` and `x=3`

Explanation:

Let's rearrange the equation a little by moving the `sqrt(2)` on the other side of the equality.
We can rewrite the equation as such:
`sqrt(2)/2 = sin(pi/2 x + (5pi)/4)`

If you remember your sine function inside the unit circle, you will notice that this value only happens when the angle is at 45 degree or 135 degree (=45˚+90˚).
In radians, these angles are `pi/4` and `3pi/4`.
More strictly speaking, this occurs everytime the angle is either
`pi/4 +2k pi` or `3pi/4 + 2k pi` where `k` is an integer.
That is, for positive integers, these angles are `pi/4`, `3pi/4`, `9pi/4`, `11pi/4`, `17pi/4`, `19pi/4`, and so on.

So the problem now reduces to solving the following:
`pi/2 x + (5pi)/4 = pi/4` (eq.1)
or
`pi/2 x + (5pi)/4 = 3pi/4` (eq.2)

(eq.1) becomes:
`pi/2 x = -pi` so `x=-2` which is not within the range [0,6),
so this cannot be a solution.

(eq.2) becomes:
`pi/2 x = -pi/2` so `x=-1` which is not within the range [0,6), so this cannot be a solution either.

Let's try some more:
`pi/2 x + (5pi)/4 = 9pi/4` (eq.3)
becomes
`pi/2 x = pi` so `x=2` and is a solution.

`pi/2 x + (5pi)/4 = 11pi/4` (eq.4)
becomes
`pi/2 x = 6pi/4` so `x=3` and is a solution.

Let's also try the next ones:
`pi/2 x + (5pi)/4 = 17pi/4` (eq.5)
becomes
`pi/2 x = 3pi` so `x=6` and is NOT a solution because it is outside the allowed range [0,6) (0 is included but 6 is excluded).

Similarly
`pi/2 x + (5pi)/4 = 19pi/4` (eq.6)
becomes
`pi/2 x = 14pi/4` so `x= 7` and is outside the range.

Ultimately, the only solutions of this equation are
`x=2` and `x=3`.
Q.E.D.