if `costheta+cos2theta+cos3theta=0`, the general value of `theta` is

Options

A. `theta=2npi+-(pi)/(4)`
B. `theta=npi+(-1)^(n)(2pi)/(3)`
C. `theta=npi+(-1)^(n)(pi)/(3)`
D. `theta=2npi+-(2pi)/(3)`

Explanation

To solve the equation \( \cos \theta + \cos 2\theta + \cos 3\theta = 0 \), we can follow these steps:

### Step 1: Use the sum-to-product identities
We can group \( \cos \theta \) and \( \cos 3\theta \) together. We will use the identity for the sum of cosines:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Let \( A = 3\theta \) and \( B = \theta \):
\[ \cos \theta + \cos 3\theta = 2 \cos\left(\frac{3\theta + \theta}{2}\right) \cos\left(\frac{3\theta - \theta}{2}\right) = 2 \cos(2\theta) \cos(\theta) \]

### Step 2: Substitute back into the equation
Now we substitute this back into the original equation:
\[ 2 \cos(2\theta) \cos(\theta) + \cos(2\theta) = 0 \]
Factoring out \( \cos(2\theta) \):
\[ \cos(2\theta)(2 \cos(\theta) + 1) = 0 \]

### Step 3: Set each factor to zero
Now we set each factor equal to zero:
1. \( \cos(2\theta) = 0 \)
2. \( 2 \cos(\theta) + 1 = 0 \)

### Step 4: Solve \( \cos(2\theta) = 0 \)
The general solution for \( \cos x = 0 \) is:
\[ x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
Thus,
\[ 2\theta = \frac{\pi}{2} + n\pi \implies \theta = \frac{\pi}{4} + \frac{n\pi}{2} \]

### Step 5: Solve \( 2 \cos(\theta) + 1 = 0 \)
Solving for \( \cos(\theta) \):
\[ 2 \cos(\theta) = -1 \implies \cos(\theta) = -\frac{1}{2} \]
The general solution for \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad \theta = \frac{4\pi}{3} + 2n\pi \quad (n \in \mathbb{Z}) \]

### Step 6: Combine the solutions
The general solutions for \( \theta \) are:
1. \( \theta = \frac{\pi}{4} + \frac{n\pi}{2} \)
2. \( \theta = \frac{2\pi}{3} + 2n\pi \)
3. \( \theta = \frac{4\pi}{3} + 2n\pi \)

### Final Answer
The general values of \( \theta \) are:
\[ \theta = \frac{\pi}{4} + \frac{n\pi}{2} \quad \text{and} \quad \theta = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad \theta = \frac{4\pi}{3} + 2n\pi \]