Question

To solve 2sin(3θ)=0, we set sin(3θ)=0, finding θ=___. In the interval [0,2π), solutions are θ=0 and θ=π.

Answer 1

To solve 2sin(3θ)=0, we look for when sin(3θ) is equal to zero, leading to solutions θ=0, θ=π/3, and θ=2π/3 in the interval [0,2π).

Step-by-step explanation:

To solve the equation 2sin(3θ)=0, we indeed set sin(3θ)=0 to find the values of θ.

Since the sine function is equal to zero at multiples of π (pi), in the interval [0,2π), the possible values for 3θ are 0, π, and 2π.

This leads to θ being equal to 0, π/3, and 2π/3, respectively, upon dividing each by 3.

However, the student states that the solutions are θ=0 and θ=π, which would correlate with 3θ being equal to 0 and 3π, respectively, missing the solution for 2θ. Therefore, the correct solutions in the interval [0,2π) should be θ=0, θ=π/3, and θ=2π/3.