To solve the problem, we need to find the values of \( \theta \) in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) such that \( \theta \neq \frac{n\pi}{5} \) for \( n = 0, \pm 1, \pm 2 \), and satisfy the equations \( \tan \theta = \cot(5\theta) \) and \( \sin(2\theta) = \cos(4\theta) \).
### Step 1: Solve \( \tan \theta = \cot(5\theta) \)
We can rewrite \( \cot(5\theta) \) as \( \frac{1}{\tan(5\theta)} \). Thus, we have:
\[
\tan \theta = \frac{1}{\tan(5\theta)}
\]
This implies:
\[
\tan \theta \tan(5\theta) = 1
\]
### Step 2: Use the identity \( \tan A \tan B = 1 \)
From the identity \( \tan A \tan B = 1 \), we can write:
\[
\theta + 5\theta = \frac{\pi}{2} + k\pi \quad (k \in \mathbb{Z})
\]
This simplifies to:
\[
6\theta = \frac{\pi}{2} + k\pi
\]
Thus:
\[
\theta = \frac{\pi}{12} + \frac{k\pi}{6}
\]
### Step 3: Find values of \( \theta \) in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \)
We will check for integer values of \( k \) to find \( \theta \):
- For \( k = -1 \):
\[
\theta = \frac{\pi}{12} - \frac{\pi}{6} = \frac{\pi}{12} - \frac{2\pi}{12} = -\frac{\pi}{12}
\]
- For \( k = 0 \):
\[
\theta = \frac{\pi}{12}
\]
- For \( k = 1 \):
\[
\theta = \frac{\pi}{12} + \frac{\pi}{6} = \frac{\pi}{12} + \frac{2\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4}
\]
- For \( k = 2 \):
\[
\theta = \frac{\pi}{12} + \frac{2\pi}{6} = \frac{\pi}{12} + \frac{4\pi}{12} = \frac{5\pi}{12}
\]
Thus, the possible values of \( \theta \) from this equation are:
\[
-\frac{\pi}{12}, \frac{\pi}{12}, \frac{\pi}{4}, \frac{5\pi}{12}
\]
### Step 4: Solve \( \sin(2\theta) = \cos(4\theta) \)
Using the identity \( \sin(2\theta) = \cos(4\theta) \), we can rewrite this as:
\[
\sin(2\theta) = \sin\left(\frac{\pi}{2} - 4\theta\right)
\]
This gives us:
\[
2\theta = \frac{\pi}{2} - 4\theta + n\pi \quad (n \in \mathbb{Z})
\]
Rearranging gives:
\[
6\theta = \frac{\pi}{2} + n\pi
\]
Thus:
\[
\theta = \frac{\pi}{12} + \frac{n\pi}{6}
\]
### Step 5: Find values of \( \theta \) in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \)
Repeating the same process as before:
- For \( n = -1 \):
\[
\theta = \frac{\pi}{12} - \frac{\pi}{6} = -\frac{\pi}{12}
\]
- For \( n = 0 \):
\[
\theta = \frac{\pi}{12}
\]
- For \( n = 1 \):
\[
\theta = \frac{\pi}{12} + \frac{\pi}{6} = \frac{3\pi}{12} = \frac{\pi}{4}
\]
- For \( n = 2 \):
\[
\theta = \frac{\pi}{12} + \frac{2\pi}{6} = \frac{5\pi}{12}
\]
### Step 6: Combine results and exclude \( \theta \neq \frac{n\pi}{5} \)
The possible values of \( \theta \) are:
\[
-\frac{\pi}{12}, \frac{\pi}{12}, \frac{\pi}{4}, \frac{5\pi}{12}
\]
Now, we need to check if any of these values are equal to \( \frac{n\pi}{5} \) for \( n = 0, \pm 1, \pm 2 \):
- \( \frac{0\pi}{5} = 0 \)
- \( \frac{1\pi}{5} = \frac{\pi}{5} \)
- \( \frac{-1\pi}{5} = -\frac{\pi}{5} \)
- \( \frac{2\pi}{5} = \frac{2\pi}{5} \)
- \( \frac{-2\pi}{5} = -\frac{2\pi}{5} \)
None of the values \( -\frac{\pi}{12}, \frac{\pi}{12}, \frac{\pi}{4}, \frac{5\pi}{12} \) match \( \frac{n\pi}{5} \).
### Final Answer
Thus, the number of valid values of \( \theta \) in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) is **4**.
