Final answer:
To solve 2sin(3θ)=0, we find the zeros of the sine function, which occur at integer multiples of π. For the interval [0,2π), the solutions are θ=0, π/3, 2π/3, π, 4π/3, and 5π/3, not merely θ=0 and θ=π as initially stated.
Step-by-step explanation:
To find the solutions for the equation 2sin(3θ)=0, we first isolate sin(3θ) by dividing both sides of the equation by 2, getting us sin(3θ)=0. The sine function is equal to zero whenever its argument is an integer multiple of π. Therefore, the solutions to sin(3θ)=0 are when 3θ is an integer multiple of π, which we can express as 3θ=nπ where n is an integer.
To determine the specific solutions within the interval [0,2π), we consider the multiples of π that would result in values within this range when divided by 3. Only n=0, 1, 2, 3, 4, and 5 will satisfy this within the interval for θ. Dividing these values by 3, we find the solutions for θ to be 0, π/3, 2π/3, π, 4π/3, and 5π/3.
However, it's important to note that if we only consider θ=0 and θ=π as solutions, we would be missing additional solutions in the given interval [0,2π). Those statements seem to be typos or incorrect information. All the derived solutions 0, π/3, 2π/3, π, 4π/3, and 5π/3 are the correct values of θ that satisfy the given equation in the interval [0,2π).