The number of solution of the equation `Sigma_(r=1)^(5)cos(rx)=0` lying in `(0, pi)` is

Answer

To find the number of solutions of the equation

\[ \sum_{r=1}^{5} \cos(rx) = 0 \]

in the interval \( (0, \pi) \), we can follow these steps:

### Step 1: Expand the summation
We start by expanding the summation:

\[ \cos(x) + \cos(2x) + \cos(3x) + \cos(4x) + \cos(5x) = 0 \]

### Step 2: Use the sum of cosines formula
We can use the formula for the sum of cosines:

\[ \sum_{r=1}^{n} \cos(rx) = \frac{\sin\left(\frac{nx}{2}\right) \cos\left(\frac{(n+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \]

For our case, \( n = 5 \):

\[ \sum_{r=1}^{5} \cos(rx) = \frac{\sin\left(\frac{5x}{2}\right) \cos\left(3x\right)}{\sin\left(\frac{x}{2}\right)} = 0 \]

### Step 3: Set the equation to zero
The equation becomes:

\[ \frac{\sin\left(\frac{5x}{2}\right) \cos(3x)}{\sin\left(\frac{x}{2}\right)} = 0 \]

This implies that either:

1. \( \sin\left(\frac{5x}{2}\right) = 0 \)
2. \( \cos(3x) = 0 \)
3. \( \sin\left(\frac{x}{2}\right) \neq 0 \) (to avoid division by zero)

### Step 4: Solve \( \sin\left(\frac{5x}{2}\right) = 0 \)
The sine function is zero at integer multiples of \( \pi \):

\[ \frac{5x}{2} = n\pi \quad \Rightarrow \quad x = \frac{2n\pi}{5} \]

We need to find \( n \) such that \( x \) lies in \( (0, \pi) \):

- For \( n = 1 \): \( x = \frac{2\pi}{5} \)
- For \( n = 2 \): \( x = \frac{4\pi}{5} \)
- For \( n = 3 \): \( x = \frac{6\pi}{5} \) (not in the interval)

Thus, we have 2 solutions from this part: \( x = \frac{2\pi}{5}, \frac{4\pi}{5} \).

### Step 5: Solve \( \cos(3x) = 0 \)
The cosine function is zero at odd multiples of \( \frac{\pi}{2} \):

\[ 3x = \frac{(2m+1)\pi}{2} \quad \Rightarrow \quad x = \frac{(2m+1)\pi}{6} \]

We need to find \( m \) such that \( x \) lies in \( (0, \pi) \):

- For \( m = 0 \): \( x = \frac{\pi}{6} \)
- For \( m = 1 \): \( x = \frac{\pi}{2} \)
- For \( m = 2 \): \( x = \frac{5\pi}{6} \)
- For \( m = 3 \): \( x = \frac{7\pi}{6} \) (not in the interval)

Thus, we have 3 solutions from this part: \( x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6} \).

### Step 6: Total number of solutions
Adding the solutions from both parts, we have:

- From \( \sin\left(\frac{5x}{2}\right) = 0 \): 2 solutions
- From \( \cos(3x) = 0 \): 3 solutions

Total solutions = \( 2 + 3 = 5 \).

### Final Answer
The number of solutions of the equation \( \sum_{r=1}^{5} \cos(rx) = 0 \) lying in \( (0, \pi) \) is **5**.

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