The given curve is $y \sin x=x+y$

Differentiate both sides with respect to $x$

$\begin{array}{rcl}\frac{dy}{dx}\sin x+y\cos x& =& 1+\frac{dy}{dx}\\ \frac{dy}{dx}\left(1-\sin x\right)& =& y\cos x-1\\ \frac{dy}{dx}& =& \frac{y\cos x-1}{1-\sin x}\end{array}$

The slope of the tangent to the curve at $x=0$ is $m=\frac{dy}{dx}at x=0$, means at origin.

Hence, $m=\frac{-1+0}{1}=-1$

Equation of tangent is $y-y_1=m\left(x-x_1\right)$
$\begin{array}{rcl}y-0& =& -1\left(x-0\right)\\ x-y& =& 0\end{array}$

Therefore, the equation of tangent to the curve $y \sin x=x+y$ at $x=0$ is $\boxed{\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}}$