Given:

#sin(x+pi/4)-cos(x)=0#

Use the identity #sin(A+B) = sin(A)cos(B)+cos(A)sin(B)# where #A = x and B = pi/4#:

#sin(x)cos(pi/4)+cos(x)sin(pi/4)-cos(x) = 0#

Use the fact that #cos(pi/4)=sin(pi/4) = sqrt2/2#:

#sin(x)sqrt2/2+cos(x)sqrt2/2-cos(x) = 0#

Multiply both sides by #sqrt2#:

#sin(x)(sqrt2sqrt2)/2+cos(x)(sqrt2sqrt2)/2-sqrt2cos(x) = 0#

Simplify by observing that #(sqrt2sqrt2)/2 = 1#:

#sin(x)+cos(x)-sqrt2cos(x) = 0#

Divide both sides by #cos(x)#:

#sin(x)/cos(x)+ 1-sqrt2 = 0#

Add #sqrt2-1# to both sides:

#sin(x)/cos(x)=sqrt2 -1#

Use the identity #sin(x)/cos(x) = tan(x)#:

#tan(x)=sqrt2 -1=tan(pi/8)#

#x = pi/8#

The inverse tangent repeats at every integer multiple of #pi#:

#x = pi/8+ npi; n in ZZ#