The first solution has a direct answer, because #arc sin{sin(z)}=z# with every number #z#, so #(5pi)/4# is a solution. It´s like adding and subtracting a number, at the end you did nothing. The second solution is because the #arc# functions have mostly two results (and even more if you consider numbers bigger than #2pi# or smaller than #-2pi#). So, the #arc sin# function has solutions that are angles, because the #arc sin# function "asks" for the angle which sine is equal to it's argument (#arc sin (1) = pi/2# because #sin(pi/2)=1#).

If you know #sin(5pi/4)=(-sqrt(2)/2)#, then you need to find the angle #x# that makes #sin(x)=(-sqrt(2)/2)#, which is the same as saying #x =arc sin(-sqrt(2)/2)=arc sin(sin((5pi)/4))#. At this point, you can do it the way you want, maybe you memorize the values of the sine and cosine functions in angles of interest and you know that the answer is #x=-pi/4# or you used your calculator. Or you can do it the way i did.

I know the answer is #-pi/4# because the sine function is the proyection in the **y** axe of the angle in an circumference that has radius 1. So, by symmetry the other angle is -45 degrees (#-pi/4#). I really hope these last lines don't make you confuse, maybe this image will help illustrate what i'm trying to say:

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