How do I solve the following math equation?

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

1 Answer
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:
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.