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)##