How do I solve the following math equation?

$1 = \sqrt{2} \sin \left(\frac{\pi}{2} x + \frac{5 \pi}{4}\right)$ for [0,6)

Feb 15, 2018

$x = 2$ 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:
$\frac{\sqrt{2}}{2} = \sin \left(\frac{\pi}{2} x + \frac{5 \pi}{4}\right)$

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 $\frac{\pi}{4}$ and $3 \frac{\pi}{4}$.
More strictly speaking, this occurs everytime the angle is either
$\frac{\pi}{4} + 2 k \pi$ or $3 \frac{\pi}{4} + 2 k \pi$ where $k$ is an integer.
That is, for positive integers, these angles are $\frac{\pi}{4}$, $3 \frac{\pi}{4}$, $9 \frac{\pi}{4}$, $11 \frac{\pi}{4}$, $17 \frac{\pi}{4}$, $19 \frac{\pi}{4}$, and so on.

So the problem now reduces to solving the following:
$\frac{\pi}{2} x + \frac{5 \pi}{4} = \frac{\pi}{4}$ (eq.1)
or
$\frac{\pi}{2} x + \frac{5 \pi}{4} = 3 \frac{\pi}{4}$ (eq.2)

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

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

Let's try some more:
$\frac{\pi}{2} x + \frac{5 \pi}{4} = 9 \frac{\pi}{4}$ (eq.3)
becomes
$\frac{\pi}{2} x = \pi$ so $x = 2$ and is a solution.

$\frac{\pi}{2} x + \frac{5 \pi}{4} = 11 \frac{\pi}{4}$ (eq.4)
becomes
$\frac{\pi}{2} x = 6 \frac{\pi}{4}$ so $x = 3$ and is a solution.

Let's also try the next ones:
$\frac{\pi}{2} x + \frac{5 \pi}{4} = 17 \frac{\pi}{4}$ (eq.5)
becomes
$\frac{\pi}{2} x = 3 \pi$ 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
$\frac{\pi}{2} x + \frac{5 \pi}{4} = 19 \frac{\pi}{4}$ (eq.6)
becomes
$\frac{\pi}{2} x = 14 \frac{\pi}{4}$ so $x = 7$ and is outside the range.

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