Note that,
#sin(pi/2+x)=sin(pi/2)cos(x)+cos(pi/2)cos(x)#,
#sin(pi/2+x)=cos(x)#.
We see that,
#cos(2x)=sin(pi/2+2x)#.
Due to the periodicity of #sin(x)# we have that #sin(pi-x)=sin(x)#, and #sin(x)=sin(x+2npi)# with #n in ZZ#. (This can be seen from expanding #sin(pi-x)# using the sine addition formulae).
Define
#f(x)="arcsin"(cos(2x))#.
Then we have
#f(x)="arcsin"(sin(pi/2+2x+2npi))#.
Look at the function #"arcsin"(x)# pictured below.
graph{arcsin(x) [-10, 10, -5, 5]}
it takes in an argument between 0 and 1 and outputs the principle value of #x# in #-pi/2<=x<=pi/2# that would make #sin(x)# output that value.
So, we have that #pi/2+2x+2npi# needs to be between #-pi/2# and #pi/2#.
I.e.,
#2x+pi(2n+1/2)< pi/2#,
#2x< pi(1/2-1/2-2n)#,
#2x< -2npi#,
#x< -npi#.
And,
#2x+pi(2n+1/2)> -pi/2#,
#2x> -pi(1+2n)#,
#x> -npi-1/2pi#
As #n# is an arbitrary integer we can easily let #n=-n# and see that for #(n-1/2)pi<x<npi# that #f(x)=2x+pi/2(1-4n)#.
Notice that for #n in ZZ# as we require this leaves some portions of #f(x)# undefined.
We can use #sin(pi-x)=sin(x)# and define #f(x)# as,
#f(x) = "arcsin"(sin(pi-pi/2-2x-2npi))#,
#f(x)="arcsin"(sin(pi/2-2x-2npi))#,
#f(x)="arcsin"(sin(pi/2-2x+2npi))#.
We need to solve the same inequalities as before,
#pi/2-2x+2npi > -pi/2# gives #x< pi/2+npi#.
#pi/2-2x+2npi < pi/2# gives #x> npi#.
Then we see that for #npi < x < pi/2+npi#, #f(x)=-2x+pi/2(4n+1)#. Now #f(x)# is defined for all #x#.
Then,
#f(x) = { (2x+pi/2(1-4n), (n-1/2)pi<=x<=npi), (-2x+pi/2(1+4n), npi<=x<=(n+1/2)pi) :}# for #n in ZZ#.
