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The given equation is:

\[
\sin(2x) \cos(x) + \sin(x) = 0
\]

We need to solve this equation for \(x\) within the interval \([0, 2\pi]\).

### Step 1: Use a trigonometric identity
Recall that \(\sin(2x) = 2\sin(x)\cos(x)\). So, substituting this into the equation, we get:

\[
2\sin(x)\cos^2(x) + \sin(x) = 0
\]

### Step 2: Factor the equation
We can factor out \(\sin(x)\):

\[
\sin(x)\left(2\cos^2(x) + 1\right) = 0
\]

Now, set each factor equal to zero.

### Step 3: Solve each part

1. \(\sin(x) = 0\)

   The solutions for \(\sin(x) = 0\) in the interval \([0, 2\pi]\) are:
   \[
   x = 0, \pi, 2\pi
   \]

2. \(2\cos^2(x) + 1 = 0\)

   Rearranging gives:
   \[
   \cos^2(x) = -\frac{1}{2}
   \]
   However, \(\cos^2(x)\) cannot be negative, so there are no solutions from this part.

### Step 4: Final solution
The solutions are \(x = 0, \pi, 2\pi\).

Thus, the solutions to the equation in the interval \([0, 2\pi]\) are:

\[
x = 0, \pi, 2\pi
\]

Let me know if you would like further details or have any questions!

Here are 5 related questions to deepen your understanding:
1. How would the solution change if the interval was \([0, 4\pi]\)?
2. Can you solve for \(x\) if the equation was \(\sin(2x) + \sin(x) = 0\)?
3. What are the solutions to \(\cos(x) = 0\) in the interval \([0, 2\pi]\)?
4. What other trigonometric identities could be helpful in similar equations?
5. How would this problem differ if it included a phase shift (e.g., \(\sin(2x + \frac{\pi}{2})\cos(x)\))?

**Tip**: Always check if factoring or using trigonometric identities simplifies a trigonometric equation before solving.

Find all solutions in the interval [0, 2π] for the equation sin(2x)cos(x) + sin(x) = 0.

Learn how to solve the trigonometric equation sin(2x)cos(x) + sin(x) = 0 in the interval [0, 2π]. This problem involves key trigonometric concepts and identities, including the double angle identity for sine and factoring techniques. Suitable for students in Grades 11-12.

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