To solve the equation \( \cos x + \cos 2x + \cos 3x + \cos 4x = 0 \) for \( 0 \leq x \leq 2\pi \), we can follow these steps:
### Step 1: Group the cosines
We can group the terms in pairs:
\[
\cos x + \cos 4x + \cos 2x + \cos 3x = 0
\]
### Step 2: Use the cosine addition formula
We can apply the cosine addition formula:
\[
\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)
\]
Applying this to \( \cos x + \cos 4x \):
\[
\cos x + \cos 4x = 2 \cos\left(\frac{5x}{2}\right) \cos\left(\frac{3x}{2}\right)
\]
And for \( \cos 2x + \cos 3x \):
\[
\cos 2x + \cos 3x = 2 \cos\left(\frac{5x}{2}\right) \cos\left(\frac{x}{2}\right)
\]
### Step 3: Combine the results
Now we can combine these results:
\[
2 \cos\left(\frac{5x}{2}\right) \cos\left(\frac{3x}{2}\right) + 2 \cos\left(\frac{5x}{2}\right) \cos\left(\frac{x}{2}\right) = 0
\]
Factoring out \( 2 \cos\left(\frac{5x}{2}\right) \):
\[
2 \cos\left(\frac{5x}{2}\right) \left( \cos\left(\frac{3x}{2}\right) + \cos\left(\frac{x}{2}\right) \right) = 0
\]
### Step 4: Set each factor to zero
This gives us two equations to solve:
1. \( \cos\left(\frac{5x}{2}\right) = 0 \)
2. \( \cos\left(\frac{3x}{2}\right) + \cos\left(\frac{x}{2}\right) = 0 \)
### Step 5: Solve \( \cos\left(\frac{5x}{2}\right) = 0 \)
The cosine function is zero at:
\[
\frac{5x}{2} = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z})
\]
This gives:
\[
x = \frac{\pi}{5} + \frac{2n\pi}{5}
\]
For \( n = 0, 1, 2, 3, 4, 5 \), we find:
- \( n = 0 \): \( x = \frac{\pi}{5} \)
- \( n = 1 \): \( x = \frac{3\pi}{5} \)
- \( n = 2 \): \( x = \pi \)
- \( n = 3 \): \( x = \frac{7\pi}{5} \)
- \( n = 4 \): \( x = \frac{9\pi}{5} \)
### Step 6: Solve \( \cos\left(\frac{3x}{2}\right) + \cos\left(\frac{x}{2}\right) = 0 \)
Using the cosine addition formula again:
\[
\cos\left(\frac{3x}{2}\right) = -\cos\left(\frac{x}{2}\right)
\]
This leads to:
\[
\cos\left(\frac{3x}{2}\right) = \cos\left(\pi - \frac{x}{2}\right)
\]
Thus, we have:
\[
\frac{3x}{2} = \pi - \frac{x}{2} + 2k\pi \quad (k \in \mathbb{Z})
\]
This simplifies to:
\[
2x = \pi + 2k\pi \quad \Rightarrow \quad x = \frac{\pi}{2} + k\pi
\]
For \( k = 0, 1 \):
- \( k = 0 \): \( x = \frac{\pi}{2} \)
- \( k = 1 \): \( x = \frac{3\pi}{2} \)
### Step 7: Compile all solutions
The solutions we have found are:
- From \( \cos\left(\frac{5x}{2}\right) = 0 \): \( \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5} \)
- From \( \cos\left(\frac{3x}{2}\right) + \cos\left(\frac{x}{2}\right) = 0 \): \( \frac{\pi}{2}, \frac{3\pi}{2} \)
### Step 8: Count the unique solutions
The unique solutions in the interval \( [0, 2\pi] \) are:
- \( \frac{\pi}{5}, \frac{3\pi}{5}, \frac{\pi}{2}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5}, \frac{3\pi}{2} \)
Thus, the total number of real values of \( x \) that satisfy the equation is **7**.
