Questions: Solve the given equation for θ ∈[0,2 π] sin (2 θ)=sin θ

Solution Steps

To solve the equation \(\sin(2\theta) = \sin(\theta)\) for \(\theta \in [0, 2\pi]\), we can use trigonometric identities and properties of sine functions. We can start by using the double-angle identity for sine, \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\). Then, we set up the equation and solve for \(\theta\) by finding the values that satisfy the equation within the given interval.

1. Use the double-angle identity: \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\).
2. Set up the equation: \(2\sin(\theta)\cos(\theta) = \sin(\theta)\).
3. Factor out \(\sin(\theta)\) and solve for \(\theta\).
4. Check for solutions within the interval \([0, 2\pi]\).

We start with the given equation: \[ \sin(2\theta) = \sin(\theta) \] Using the double-angle identity for sine, we have: \[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \] Thus, the equation becomes: \[ 2\sin(\theta)\cos(\theta) = \sin(\theta) \]

We can factor out \(\sin(\theta)\) from both sides: \[ 2\sin(\theta)\cos(\theta) - \sin(\theta) = 0 \] \[ \sin(\theta)(2\cos(\theta) - 1) = 0 \] This gives us two possible solutions:

1. \(\sin(\theta) = 0\)
2. \(2\cos(\theta) - 1 = 0\)

For \(\sin(\theta) = 0\): \[ \theta = 0, \pi, 2\pi \]

For \(2\cos(\theta) - 1 = 0\): \[ \cos(\theta) = \frac{1}{2} \] \[ \theta = \frac{\pi}{3}, \frac{5\pi}{3} \]

Combining all solutions within the interval \([0, 2\pi]\), we get: \[ \theta = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi \]

Final Answer