The number of solutions of `sum_(r=1)^5cosrx=5` in the interval `[0,2pi]` is 0 (b) 2 (c) 5 (d) 10

Options

A. 0
B. 2
C. 5
D. 10

Explanation

To solve the equation \(\sum_{r=1}^{5} \cos(rx) = 5\) in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Understand the Equation
The equation \(\sum_{r=1}^{5} \cos(rx) = 5\) means that we need to find the values of \(x\) for which the sum of the cosines of multiples of \(x\) from \(1\) to \(5\) equals \(5\).

### Step 2: Analyze the Range of Cosine
The cosine function, \(\cos(rx)\), has a maximum value of \(1\). Therefore, for the sum of five cosine terms to equal \(5\), each 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: Find the Values of \(x\)
The cosine function equals \(1\) at:
\[ \cos(k) = 1 \implies k = 2n\pi \quad (n \in \mathbb{Z}) \]
Thus, for each \(r\):
\[ rx = 2n\pi \implies x = \frac{2n\pi}{r} \]

### Step 4: Determine Values of \(x\) for Each \(r\)
Now, we will find the values of \(x\) for \(r = 1, 2, 3, 4, 5\):
- For \(r = 1\): \(x = 2n\pi\)
- For \(r = 2\): \(x = \frac{2n\pi}{2} = n\pi\)
- For \(r = 3\): \(x = \frac{2n\pi}{3}\)
- For \(r = 4\): \(x = \frac{2n\pi}{4} = \frac{n\pi}{2}\)
- For \(r = 5\): \(x = \frac{2n\pi}{5}\)

### Step 5: Find Valid Solutions in the Interval \([0, 2\pi]\)
Now we will find the valid solutions for \(n = 0\) and \(n = 1\) (since \(n = 2\) gives values outside \([0, 2\pi]\)):
- For \(n = 0\): \(x = 0\)
- For \(n = 1\):
- \(r = 1\): \(x = 2\pi\)
- \(r = 2\): \(x = \pi\)
- \(r = 3\): \(x = \frac{2\pi}{3}\)
- \(r = 4\): \(x = \frac{\pi}{2}\)
- \(r = 5\): \(x = \frac{2\pi}{5}\)

### Step 6: List All Unique Solutions
The unique solutions in the interval \([0, 2\pi]\) are:
- \(0\)
- \(\frac{2\pi}{5}\)
- \(\frac{\pi}{2}\)
- \(\pi\)
- \(2\pi\)

### Conclusion
Thus, the total number of solutions in the interval \([0, 2\pi]\) is \(5\).

The answer is **(c) 5**.