Final answer:
The solutions to 2sin(3θ) = 0 are found by considering the points where sine is zero. In the interval (0, 2π), the correct solutions are θ = 0, π/3, π, 4π/3, and 5π/3, excluding 2π/3 which was mentioned by the student. Additionally, for wave equations, the presence of nodes or antinodes depends on the values of sine and cosine functions at given positions and times.
Step-by-step explanation:
The solutions to the equation (2sin(3θ) = 0) are found by setting the sine term equal to zero and solving for θ. As sine is zero at integer multiples of π, the solutions are θ = nπ/3, where n is an integer. But to ensure the solutions are in the interval (0, 2π), we find that the correct solutions are θ = 0, π/3, 2π/3, π, 4π/3, and 5π/3. The students response seems to have misunderstood the solutions in the interval, as θ should not include 2π/3 for the interval (0, 2).
When we look at the equation y (x, t) = 4.00 cm sin (3 m⁻¹x + 2) cos (4 s⁻¹t), we see that there's an antinode at x = 0 because the sine term equals 1. The cosine term causes the system to oscillate between its maximum amplitude, A, and its negative maximum, -A, which represents the crystal clear movement back and forth from the maximum displacement in a wave.