To find the number of solutions to the equation
\[
\sum_{r=1}^{5} \cos(rx) = 5
\]
in the interval \([0, 4\pi]\), we can follow these steps:
### Step 1: Understanding the Equation
The equation can be expanded as:
\[
\cos(x) + \cos(2x) + \cos(3x) + \cos(4x) + \cos(5x) = 5
\]
### Step 2: Analyzing the Range of Cosine Functions
The cosine function has a maximum value of 1. Therefore, each term \(\cos(rx)\) (for \(r = 1, 2, 3, 4, 5\)) can contribute at most 1 to the sum.
Since there are 5 terms, the maximum possible value of the sum is:
\[
1 + 1 + 1 + 1 + 1 = 5
\]
This means that for the equation to hold true, each cosine term must equal 1:
\[
\cos(x) = 1, \quad \cos(2x) = 1, \quad \cos(3x) = 1, \quad \cos(4x) = 1, \quad \cos(5x) = 1
\]
### Step 3: Finding the General Solutions
The cosine function equals 1 at:
\[
\theta = 2n\pi \quad \text{for } n \in \mathbb{Z}
\]
Thus, we can write the conditions for each term:
1. \(x = 2n\pi\)
2. \(2x = 2n\pi \Rightarrow x = n\pi\)
3. \(3x = 2n\pi \Rightarrow x = \frac{2n\pi}{3}\)
4. \(4x = 2n\pi \Rightarrow x = \frac{n\pi}{2}\)
5. \(5x = 2n\pi \Rightarrow x = \frac{2n\pi}{5}\)
### Step 4: Finding Solutions in the Interval \([0, 4\pi]\)
Now we will find the values of \(x\) for each case within the interval \([0, 4\pi]\):
1. From \(x = 2n\pi\):
- \(n = 0 \Rightarrow x = 0\)
- \(n = 1 \Rightarrow x = 2\pi\)
- \(n = 2 \Rightarrow x = 4\pi\)
2. From \(x = n\pi\):
- \(n = 0 \Rightarrow x = 0\)
- \(n = 1 \Rightarrow x = \pi\)
- \(n = 2 \Rightarrow x = 2\pi\)
- \(n = 3 \Rightarrow x = 3\pi\)
- \(n = 4 \Rightarrow x = 4\pi\)
3. From \(x = \frac{2n\pi}{3}\):
- \(n = 0 \Rightarrow x = 0\)
- \(n = 1 \Rightarrow x = \frac{2\pi}{3}\)
- \(n = 2 \Rightarrow x = \frac{4\pi}{3}\)
- \(n = 3 \Rightarrow x = 2\pi\)
- \(n = 4 \Rightarrow x = \frac{8\pi}{3}\)
- \(n = 5 \Rightarrow x = \frac{10\pi}{3}\)
4. From \(x = \frac{n\pi}{2}\):
- \(n = 0 \Rightarrow x = 0\)
- \(n = 1 \Rightarrow x = \frac{\pi}{2}\)
- \(n = 2 \Rightarrow x = \pi\)
- \(n = 3 \Rightarrow x = \frac{3\pi}{2}\)
- \(n = 4 \Rightarrow x = 2\pi\)
- \(n = 5 \Rightarrow x = \frac{5\pi}{2}\)
- \(n = 6 \Rightarrow x = 3\pi\)
- \(n = 7 \Rightarrow x = \frac{7\pi}{2}\)
- \(n = 8 \Rightarrow x = 4\pi\)
5. From \(x = \frac{2n\pi}{5}\):
- \(n = 0 \Rightarrow x = 0\)
- \(n = 1 \Rightarrow x = \frac{2\pi}{5}\)
- \(n = 2 \Rightarrow x = \frac{4\pi}{5}\)
- \(n = 3 \Rightarrow x = \frac{6\pi}{5}\)
- \(n = 4 \Rightarrow x = \frac{8\pi}{5}\)
- \(n = 5 \Rightarrow x = 2\pi\)
- \(n = 6 \Rightarrow x = \frac{12\pi}{5}\)
- \(n = 7 \Rightarrow x = \frac{14\pi}{5}\)
- \(n = 8 \Rightarrow x = \frac{16\pi}{5}\)
- \(n = 9 \Rightarrow x = \frac{18\pi}{5}\)
- \(n = 10 \Rightarrow x = 4\pi\)
### Step 5: Counting Unique Solutions
Now we will count the unique solutions from all the cases above within the interval \([0, 4\pi]\):
- From \(x = 2n\pi\): \(0, 2\pi, 4\pi\) (3 solutions)
- From \(x = n\pi\): \(0, \pi, 2\pi, 3\pi, 4\pi\) (5 solutions, but \(0, 2\pi, 4\pi\) are duplicates)
- From \(x = \frac{2n\pi}{3}\): \(0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi, \frac{8\pi}{3}, \frac{10\pi}{3}\) (6 solutions, but \(0, 2\pi\) are duplicates)
- From \(x = \frac{n\pi}{2}\): \(0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \frac{5\pi}{2}, 3\pi, \frac{7\pi}{2}, 4\pi\) (9 solutions, but \(0, 2\pi, 4\pi\) are duplicates)
- From \(x = \frac{2n\pi}{5}\): \(0, \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, \frac{8\pi}{5}, 2\pi, \frac{12\pi}{5}, \frac{14\pi}{5}, \frac{16\pi}{5}, \frac{18\pi}{5}, 4\pi\) (11 solutions, but \(0, 2\pi, 4\pi\) are duplicates)
### Final Count of Unique Solutions
Collecting all unique solutions, we find:
- \(0, \frac{\pi}{2}, \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{6\pi}{5}, \frac{8\pi}{5}, \frac{8\pi}{3}, \frac{10\pi}{3}, \frac{3\pi}{2}, 2\pi, 3\pi, \frac{5\pi}{2}, \frac{7\pi}{2}, 4\pi\)
Counting these gives us a total of 7 unique solutions.
### Conclusion
The number of solutions of the equation \(\sum_{r=1}^{5} \cos(rx) = 5\) in the interval \([0, 4\pi]\) is:
\[
\boxed{7}
\]
