If `0 le x lt 2pi`, then the number of real values of x which satisfy the equation `cosx+cos2x+cos3x+cos4x=0`, is :

Options

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

Answer

To solve the equation \( \cos x + \cos 2x + \cos 3x + \cos 4x = 0 \) for \( 0 \leq x < 2\pi \), we can follow these steps:

### Step 1: Grouping the Terms
We can group the terms in pairs:
\[ (\cos x + \cos 3x) + (\cos 2x + \cos 4x) = 0 \]

### Step 2: Applying the Cosine Addition Formula
Using the identity \( \cos a + \cos b = 2 \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \), we can rewrite the grouped terms:
\[ \cos x + \cos 3x = 2 \cos\left(2x\right) \cos\left(x\right) \]
\[ \cos 2x + \cos 4x = 2 \cos\left(3x\right) \cos\left(x\right) \]

Thus, we can rewrite the equation as:
\[ 2 \cos(2x) \cos(x) + 2 \cos(3x) \cos(x) = 0 \]

### Step 3: Factoring Out Common Terms
Factoring out \( 2 \cos(x) \):
\[ 2 \cos(x) \left( \cos(2x) + \cos(3x) \right) = 0 \]

### Step 4: Setting Each Factor to Zero
This gives us two equations to solve:
1. \( \cos(x) = 0 \)
2. \( \cos(2x) + \cos(3x) = 0 \)

### Step 5: Solving \( \cos(x) = 0 \)
The solutions for \( \cos(x) = 0 \) in the interval \( 0 \leq x < 2\pi \) are:
\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \]
This gives us **2 solutions**.

### Step 6: Solving \( \cos(2x) + \cos(3x) = 0 \)
Using the identity again:
\[ \cos(2x) + \cos(3x) = 2 \cos\left(\frac{5x}{2}\right) \cos\left(\frac{x}{2}\right) = 0 \]

This gives us two more equations:
1. \( \cos\left(\frac{5x}{2}\right) = 0 \)
2. \( \cos\left(\frac{x}{2}\right) = 0 \)

### Step 7: Solving \( \cos\left(\frac{5x}{2}\right) = 0 \)
The solutions for \( \cos\left(\frac{5x}{2}\right) = 0 \) are:
\[ \frac{5x}{2} = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
This leads to:
\[ x = \frac{\pi}{5} + \frac{2n\pi}{5} \]
Finding values for \( n \) such that \( 0 \leq x < 2\pi \):
- For \( n = 0 \): \( x = \frac{\pi}{5} \)
- For \( n = 1 \): \( x = \frac{3\pi}{5} \)
- For \( n = 2 \): \( x = \pi \)
- For \( n = 3 \): \( x = \frac{7\pi}{5} \)
- For \( n = 4 \): \( x = \frac{9\pi}{5} \)

This gives us **5 solutions**.

### Step 8: Solving \( \cos\left(\frac{x}{2}\right) = 0 \)
The solutions for \( \cos\left(\frac{x}{2}\right) = 0 \) are:
\[ \frac{x}{2} = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
This leads to:
\[ x = \pi + 2n\pi \]
Finding values for \( n \) such that \( 0 \leq x < 2\pi \):
- For \( n = 0 \): \( x = \pi \)

This gives us **1 additional solution**.

### Final Count of Solutions
Now, we combine all the solutions:
- From \( \cos(x) = 0 \): 2 solutions
- From \( \cos(2x) + \cos(3x) = 0 \): 5 solutions
- From \( \cos\left(\frac{x}{2}\right) = 0 \): 1 solution

However, \( x = \pi \) is counted in both the second and third equations, so we need to avoid double counting.

Thus, the total number of unique solutions is:
\[ 2 + 5 + 1 - 1 = 7 \]

### Conclusion
The number of real values of \( x \) that satisfy the equation \( \cos x + \cos 2x + \cos 3x + \cos 4x = 0 \) in the interval \( 0 \leq x < 2\pi \) is **7**.

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