How do I solve the following math equation?

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

1 Answer
Feb 15, 2018

Answer:

#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.