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The equation to solve is:

\[
4 \sin^5(\theta) + \sin(5\theta) = 0
\]

We are tasked with finding all solutions of this equation in the interval \( 0 \leq \theta \leq 2\pi \).

### Step 1: Factorization
We start by trying to factor the equation. Notice that both terms involve sine functions, and we can factor out a \( \sin(\theta) \) from the first term:

\[
4 \sin^5(\theta) + \sin(5\theta) = \sin(\theta)(4 \sin^4(\theta) + \sin(5\theta)) = 0
\]

This gives us two cases to solve:

1. \( \sin(\theta) = 0 \)
2. \( 4 \sin^4(\theta) + \sin(5\theta) = 0 \)

### Step 2: Solving \( \sin(\theta) = 0 \)
The sine function is zero at:

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

These are three solutions: \( \theta = 0, \pi, 2\pi \).

### Step 3: Solving \( 4 \sin^4(\theta) + \sin(5\theta) = 0 \)
This equation is more complex, involving a quintic sine term. However, we can use known solutions for \( \sin(5\theta) \) by considering angles at which \( \sin(5\theta) = 0 \), because these are multiples of \( \pi \). These occur at:

\[
5\theta = n\pi \quad \Rightarrow \quad \theta = \frac{n\pi}{5}
\]

where \( n \) is an integer. Now we find all values of \( \theta \) in the range \( 0 \leq \theta \leq 2\pi \):

- For \( n = 0 \), \( \theta = 0 \)
- For \( n = 1 \), \( \theta = \frac{\pi}{5} \)
- For \( n = 2 \), \( \theta = \frac{2\pi}{5} \)
- For \( n = 3 \), \( \theta = \frac{3\pi}{5} \)
- For \( n = 4 \), \( \theta = \frac{4\pi}{5} \)
- For \( n = 5 \), \( \theta = \pi \)
- For \( n = 6 \), \( \theta = \frac{6\pi}{5} \)
- For \( n = 7 \), \( \theta = \frac{7\pi}{5} \)
- For \( n = 8 \), \( \theta = \frac{8\pi}{5} \)
- For \( n = 9 \), \( \theta = \frac{9\pi}{5} \)
- For \( n = 10 \), \( \theta = 2\pi \)

### Step 4: Combine Solutions
The solutions to the equation are all the distinct values of \( \theta \) found from both cases. These are:

\[
\theta = 0, \frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}, \pi, \frac{6\pi}{5}, \frac{7\pi}{5}, \frac{8\pi}{5}, \frac{9\pi}{5}, 2\pi
\]

### Final Answer:
The solutions are:

\[
\theta = 0, \frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}, \pi, \frac{6\pi}{5}, \frac{7\pi}{5}, \frac{8\pi}{5}, \frac{9\pi}{5}, 2\pi
\]

---

Would you like further details on any part of this solution? Let me know if you have questions or would like additional clarifications.

Here are some follow-up questions to deepen your understanding:
1. Why does factoring \( \sin(\theta) \) help simplify trigonometric equations?
2. What are some common methods for solving quintic trigonometric equations?
3. How would the solution change if the interval was different, say \( 0 \leq \theta \leq \pi \)?
4. What is the significance of the periodicity of the sine function in solving these equations?
5. Can you apply this method to other functions involving higher powers of sine or cosine?

**Tip:** When solving trigonometric equations, always check for ways to factor the expression first, as this often leads to simpler sub-problems.

Find all the solutions of the equation 4sin^5(θ) + sin(5θ) = 0 which lie in the interval 0 ≤ θ ≤ 2π.

Learn how to solve the trigonometric equation 4sin^5(θ) + sin(5θ) = 0 over the interval 0 ≤ θ ≤ 2π. This solution covers factoring trigonometric equations, using multiple angle identities, and finding the zeros of the sine function.

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