Answer
$$ y(x)=-\cos x+ \sin x$$
Work Step by Step
Given$$(\sin x) y^{\prime}=(\cos x) y+1, \quad y\left(\frac{\pi}{4}\right)=0$$
Then $$y^{\prime}-(\cot x) y=\csc x$$
This is a linear equation with $p(x) =-\cot x,\ \ q(x) =\csc x$, so
\begin{align*}
\mu(x)&=e^{\int p(x)dx}\\
&=e^{\int -\cot x dx}\\
&=e^{-\ln \sin x}\\
&=e^{\ln \csc x}\\
&=\csc x
\end{align*}
Then
\begin{align*}
y\mu(x) &=\int \mu(x)q(x)dx\\
\csc xy &=\int \csc^2 xdx\\
&=-\cot x+C
\end{align*}
Then $$(\csc x) y=-\cot x+C\ \ \to \ \ \ y(x)=-\cos x+c \sin x$$
Since $y(\pi/4)=0 $, then $C=1$, and hence
$$ y(x)=-\cos x+ \sin x$$
