What is the domain of the function \(f(x)=\sqrt{\sin x}\) when \(x\in[0,2\pi]\)?

Correct answer: \([0,\pi]\)

For a real-valued square root, we require \(\sin x\ge 0\). Within \([0,2\pi]\), sine is non-negative on \([0,\pi]\). The endpoints are included because \(\sin 0=\sin\pi=0\). Exam tip: for a square-root function, first impose that the expression inside the root is at least zero, then intersect the result with the given interval.