Find all values of x such that 
sin(2x) = sin(x)
  and  
0 ≤ x ≤ 2𝜋.
 (Enter your answers as a comma-separated list.)


To solve the equation \( \sin(2x) = \sin(x) \) within the interval \( 0 \leq x \leq 2\pi \), we can use trigonometric identities and properties of sine functions.

### Step 1: Use the identity for \( \sin(2x) \)
We know that:
\[
\sin(2x) = 2\sin(x)\cos(x)
\]
So, the equation becomes:
\[
2\sin(x)\cos(x) = \sin(x)
\]

### Step 2: Factor the equation
We can factor out \( \sin(x) \) from both sides:
\[
\sin(x) \cdot (2\cos(x) - 1) = 0
\]

This equation will be satisfied if either:
\[
\sin(x) = 0
\]
or
\[
2\cos(x) - 1 = 0
\]

### Step 3: Solve each equation separately

**Case 1:** \( \sin(x) = 0 \)

The sine function equals zero at:
\[
x = 0, \pi, 2\pi
\]

**Case 2:** \( 2\cos(x) - 1 = 0 \)

Solve for \( \cos(x) \):
\[
\cos(x) = \frac{1}{2}
\]
The cosine function equals \( \frac{1}{2} \) at:
\[
x = \frac{\pi}{3}, \frac{5\pi}{3}
\]

### Step 4: Combine all solutions
The values of \( x \) that satisfy the original equation are:
\[
x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi
\]

Thus, the solution set is:
\[
\boxed{0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi}
\]

Would you like further details on how these solutions were found or have any other questions?

Here are five related questions you might find interesting:
1. How do we solve trigonometric equations involving tangent?
2. What are the general solutions for the equation \( \sin(x) = k \)?
3. How does the double-angle identity for sine function work in other equations?
4. How can we solve \( \cos(2x) = \cos(x) \) for \( 0 \leq x \leq 2\pi \)?
5. What are the zeros of the sine function on a given interval?

**Tip:** When solving trigonometric equations, always check if you can factor the equation to make it simpler to solve.

Learn how to solve the equation sin(2x) = sin(x) within the interval [0, 2π]. This problem involves trigonometric identities and properties of the sine function. Suitable for high school students.

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