What is the period of the function \(\sin 3x\)?

Correct answer: \(\dfrac{2\pi}{3}\)

The governing transformation rule is that the period of \(\sin(kx)\), for nonzero k, is \(2\pi/|k|\). To verify it here, require \(\sin[3(x+T)]=\sin(3x)\). The sine argument must increase by one full cycle, so \(3T=2\pi\), giving \(T=2\pi/3\). Thus option C is correct. The coefficient 3 compresses the graph horizontally by a factor of 3, so its period is one-third of the basic sine period \(2\pi\). Option A ignores the coefficient, option B is only half of the required period, and option D corresponds to \(\pi/3\), which does not generally reproduce the same sine value.