Solving sin(2θ) = sin(θ) involves using the double-angle identity, resulting in θ = nπ, π/3 + 2nπ, and 5π/3 + 2nπ, where n is an integer.
To solve the trigonometric equation sin(2θ) = sin(θ), we can apply trigonometric identities to simplify and find the solutions.
Firstly, we use the double-angle identity for sine, which states that sin(2θ) = 2sin(θ)cos(θ). Substituting this into the equation gives 2sin(θ)cos(θ) = sin(θ).
Next, we can rearrange the equation by subtracting sin(θ) from both sides, resulting in 2sin(θ)cos(θ) - sin(θ) = 0. Factoring out sin(θ), we get sin(θ)(2cos(θ) - 1) = 0.
This equation implies that either sin(θ) = 0 or (2cos(θ) - 1) = 0.
For sin(θ) = 0, θ can be any integer multiple of π, as sin(θ) is 0 at these points.
For 2cos(θ) - 1 = 0, we find cos(θ) = 1/2. This occurs at π/3 and 5π/3.
Thus, the solutions for θ are θ = nπ, π/3 + 2nπ, and 5π/3 + 2nπ, where n is an integer.