How do I solve the initial-value problem: ##y’=sinx/siny;y(0)=π/4##

First of all, let’s write ##y’## as ##dy/dx##. The expression becomes
##dy/dx = sin(x)/sin(y)##
Multiply both sides for ##sin(y) dx## and obtain
##sin(y) dy = sin(x) dx##
integrating, one has
##cos(y) = cos(x)+ c##
and thus
##y = cos^{ -1}(cos(x)+c)##
This is the general solution of the problem, and we can fix the constant ##c##, given the condition ##y(0)=pi/4##. In fact,
##y(0)=cos^{ -1}(cos(0)+c) = cos^{ -1}(1+c)=pi/4##
which means
##1+c=1/sqrt{2}##, and finally
##c=1/sqrt{2} -1##.
Your solution is thus
##y = cos^{ -1}(cos(x)+1/sqrt{2} -1)##