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 \( -1 \leq \sin(x) \leq 1 \), and for \( \frac{1}{x-3} \), the domain is any real number except \( x \neq 3 \).

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 \( -1 \leq \sin(x) \leq 1 \) and \( x \neq 3 \).

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 \( \sin(x) \geq 0 \). 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.