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The given equation is:

\[
\sin(4\theta) = \cos(5\theta)
\]

To solve this, we can use the identity that connects sine and cosine:

\[
\cos x = \sin \left( \frac{\pi}{2} - x \right)
\]

Using this identity, we can rewrite \(\cos(5\theta)\) as:

\[
\cos(5\theta) = \sin\left(\frac{\pi}{2} - 5\theta\right)
\]

Therefore, the equation becomes:

\[
\sin(4\theta) = \sin\left(\frac{\pi}{2} - 5\theta\right)
\]

Now, for two sine functions to be equal, their arguments must either be equal or differ by an integer multiple of \(2\pi\). This gives us two cases:

### Case 1:
\[
4\theta = \frac{\pi}{2} - 5\theta + 2n\pi
\]
Solving for \(\theta\):
\[
9\theta = \frac{\pi}{2} + 2n\pi
\]
\[
\theta = \frac{\pi}{18} + \frac{2n\pi}{9}
\]

### Case 2:
\[
4\theta = \pi - \left( \frac{\pi}{2} - 5\theta \right) + 2n\pi
\]
Simplifying:
\[
4\theta = \pi - \frac{\pi}{2} + 5\theta + 2n\pi
\]
\[
4\theta = \frac{\pi}{2} + 5\theta + 2n\pi
\]
Rearranging to solve for \(\theta\):
\[
- \theta = \frac{\pi}{2} + 2n\pi
\]
\[
\theta = - \left( \frac{\pi}{2} + 2n\pi \right)
\]

These are the general solutions. If you need further explanation or specific values for \(\theta\), let me know!

Would you like more details or have any questions?  
Here are 5 related questions to explore further:

1. How can we derive the sine and cosine identities used in this solution?
2. What are the general solutions for similar equations involving trigonometric functions?
3. How do we solve trigonometric equations involving multiple angles?
4. What are the methods to verify the solutions of trigonometric equations?
5. How can graphing help in visualizing the solutions of \(\sin(4\theta) = \cos(5\theta)\)?

**Tip:** For trigonometric equations, using identities to transform one function into another can simplify the solving process significantly.

Find the solution for Sin(4θ) = Cos(5θ)

This solution solves the trigonometric equation Sin(4θ) = Cos(5θ) using sine and cosine identities. Learn how to derive and simplify the general solutions for θ, covering essential trigonometric principles suitable for high school students.

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