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

Correct answer: \( \left[0,\frac{\pi}{2}\right]\cup\left[\frac{3\pi}{2},2\pi\right] \)

The direct answer is A: the domain is \\( [0,\pi/2]\cup[3\pi/2,2\pi] \\). For a real square root, the quantity inside must satisfy \\(\cos x\ge0\\). On the unit circle, cosine is the x-coordinate: it is non-negative in the first and fourth quadrants. Within \\([0,2\pi]\\), these are \\([0,\pi/2]\\) and \\([3\pi/2,2\pi]\\), including endpoints where cosine is 0. A is correct. B describes the second and third quadrants, where cosine is non-positive and is negative except at endpoints, so the square root is generally not real. C includes [0,π], but cosine is negative between π/2 and π. D includes the entire interval and therefore includes points such as π where \\(\cos\pi=-1\\). Memory cue: for \\(\sqrt{\cos x}\\), keep the right half of the unit circle where cosine is non-negative.