Find the domain of the function $y=\sqrt{\sin (x)}+\frac{1}{x-3}$.

Step 1 ：Step1: Identify the constraints of each function. For $\sqrt{\sin (x)}$, the domain is any real number such that $-1\le \sin (x)\le 1$, and for $\frac{1}{x-3}$, the domain is any real number except $x\ne 3$.

Step 2 ：Step2: Since the domains must satisfy both functions, the overall domain is the intersection of the two domains. So, we have to find the x-values that satisfy both $-1\le \sin (x)\le 1$ and $x\ne 3$.

Step 3 ：Step3: Since the range of $\sin (x)$ is $[-1,1]$, the square root function $\sqrt{\sin (x)}$ is defined for all x-values where $\sin (x)\ge 0$. This occurs when $x$ is in the interval $[0,\pi ]$ in each period of $2\pi$.

Step 4 ：Step4: Hence, the domain of $y=\sqrt{\sin (x)}+\frac{1}{x-3}$ is any x-value in the interval $[0,\pi ]$ in each period of $2\pi$, except for $x=3$, if $3$ is in the interval for a particular period.