The number of roots of equation `cos x + cos 2x + cos 3x = 0` is `( 0 le x le 2 pi)`

Options

A. 4
B. 5
C. 6
D. 8

Answer

To find the number of roots of the equation \( \cos x + \cos 2x + \cos 3x = 0 \) in the interval \( [0, 2\pi] \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos x + \cos 2x + \cos 3x = 0 \]

### Step 2: Use the cosine addition formula
We can use the cosine addition formula to combine the terms. Recall that:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
We can group \( \cos x \) and \( \cos 3x \):
\[ \cos x + \cos 3x = 2 \cos\left(\frac{x + 3x}{2}\right) \cos\left(\frac{x - 3x}{2}\right) = 2 \cos(2x) \cos(x) \]
Thus, our equation becomes:
\[ 2 \cos(2x) \cos(x) + \cos(2x) = 0 \]

### Step 3: Factor the equation
Now we can factor out \( \cos(2x) \):
\[ \cos(2x)(2 \cos(x) + 1) = 0 \]

### Step 4: Set each factor to zero
Now we set each factor to zero:
1. \( \cos(2x) = 0 \)
2. \( 2 \cos(x) + 1 = 0 \)

### Step 5: Solve \( \cos(2x) = 0 \)
The equation \( \cos(2x) = 0 \) gives us:
\[ 2x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
Thus,
\[ x = \frac{\pi}{4} + \frac{n\pi}{2} \]
Now we find the values of \( x \) in the interval \( [0, 2\pi] \):
- For \( n = 0 \): \( x = \frac{\pi}{4} \)
- For \( n = 1 \): \( x = \frac{3\pi}{4} \)
- For \( n = 2 \): \( x = \frac{5\pi}{4} \)
- For \( n = 3 \): \( x = \frac{7\pi}{4} \)

This gives us 4 solutions from \( \cos(2x) = 0 \).

### Step 6: Solve \( 2 \cos(x) + 1 = 0 \)
The equation \( 2 \cos(x) + 1 = 0 \) simplifies to:
\[ \cos(x) = -\frac{1}{2} \]
The solutions for \( \cos(x) = -\frac{1}{2} \) in the interval \( [0, 2\pi] \) are:
\[ x = \frac{2\pi}{3}, \quad x = \frac{4\pi}{3} \]
This gives us 2 additional solutions.

### Step 7: Count the total number of solutions
Adding the solutions from both parts:
- From \( \cos(2x) = 0 \): 4 solutions
- From \( 2 \cos(x) + 1 = 0 \): 2 solutions

Thus, the total number of roots in the interval \( [0, 2\pi] \) is:
\[ 4 + 2 = 6 \]

### Final Answer
The number of roots of the equation \( \cos x + \cos 2x + \cos 3x = 0 \) in the interval \( [0, 2\pi] \) is **6**.

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