If `0 le x le 2pi`, then the number of real values of x, which satisfy the equation `cos x + cos 2x + cos 3x + cos 4x=0`, is

Options

A. 3
B. 5
C. 7
D. 9

Explanation

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 Cosine Terms
We can group the terms in pairs:
\[ \cos x + \cos 4x + \cos 2x + \cos 3x = 0 \]

### Step 2: Use the Cosine Addition Formula
Using the cosine addition formula, we can rewrite the pairs:
\[ \cos x + \cos 4x = 2 \cos\left(\frac{5x}{2}\right) \cos\left(\frac{3x}{2}\right) \]
\[ \cos 2x + \cos 3x = 2 \cos\left(\frac{5x}{2}\right) \cos\left(\frac{x}{2}\right) \]

### Step 3: Combine the Results
Now, we can combine these results:
\[ 2 \cos\left(\frac{5x}{2}\right) \left( \cos\left(\frac{3x}{2}\right) + \cos\left(\frac{x}{2}\right) \right) = 0 \]

### Step 4: Set Each Factor to Zero
This gives us two cases 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 the First Case
For \( \cos\left(\frac{5x}{2}\right) = 0 \):
\[ \frac{5x}{2} = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
\[ x = \frac{2}{5} \left(\frac{\pi}{2} + n\pi\right) = \frac{\pi}{5} + \frac{2n\pi}{5} \]
We need to find values of \( n \) such that \( 0 \leq x \leq 2\pi \):
- For \( n = 0 \): \( x = \frac{\pi}{5} \)
- For \( n = 1 \): \( x = \frac{7\pi}{5} \)
- For \( n = 2 \): \( x = \frac{13\pi}{5} \) (not valid as it exceeds \( 2\pi \))

Thus, from this case, we have two solutions: \( x = \frac{\pi}{5}, \frac{7\pi}{5} \).

### Step 6: Solve the Second Case
For \( \cos\left(\frac{3x}{2}\right) + \cos\left(\frac{x}{2}\right) = 0 \):
Using the identity \( \cos A + \cos B = 0 \):
\[ \cos\left(\frac{3x}{2}\right) = -\cos\left(\frac{x}{2}\right) \]
This implies:
\[ \frac{3x}{2} = \pi + \frac{x}{2} + 2k\pi \quad (k \in \mathbb{Z}) \]
Solving for \( x \):
\[ \frac{3x}{2} - \frac{x}{2} = \pi + 2k\pi \]
\[ x = \pi + 2k\pi \]
Valid values of \( k \) for \( 0 \leq x \leq 2\pi \):
- For \( k = 0 \): \( x = \pi \)
- For \( k = 1 \): \( x = 3\pi \) (not valid)

Thus, from this case, we have one solution: \( x = \pi \).

### Step 7: Combine All Solutions
Combining all the solutions we found:
- From the first case: \( x = \frac{\pi}{5}, \frac{7\pi}{5} \)
- From the second case: \( x = \pi \)

Thus, the total number of solutions is:
\[ \frac{\pi}{5}, \pi, \frac{7\pi}{5} \]
This gives us a total of **three distinct solutions**.

### Final Answer
The number of real values of \( x \) that satisfy the equation \( \cos x + \cos 2x + \cos 3x + \cos 4x = 0 \) in the interval \( 0 \leq x \leq 2\pi \) is **3**.