If `costheta+cos3theta+cos5theta+cos7theta=0`, then general value of `theta` is:

Options

A. `(2n+1)pi/2`
B. `(2n+1)pi/3`
C. `(2n+1)pi/4`
D. `(2n+1)pi/6`

Answer

To solve the equation \( \cos \theta + \cos 3\theta + \cos 5\theta + \cos 7\theta = 0 \), we will follow these steps:

### Step 1: Grouping the Terms
We can group the terms as follows:
\[ (\cos \theta + \cos 7\theta) + (\cos 3\theta + \cos 5\theta) = 0 \]

### 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 apply it to each group.

For the first group \( \cos \theta + \cos 7\theta \):
- Here, \( A = \theta \) and \( B = 7\theta \).
- Thus, we have:
\[ \cos \theta + \cos 7\theta = 2 \cos \left( \frac{\theta + 7\theta}{2} \right) \cos \left( \frac{\theta - 7\theta}{2} \right) = 2 \cos(4\theta) \cos(3\theta) \]

For the second group \( \cos 3\theta + \cos 5\theta \):
- Here, \( A = 3\theta \) and \( B = 5\theta \).
- Thus, we have:
\[ \cos 3\theta + \cos 5\theta = 2 \cos \left( \frac{3\theta + 5\theta}{2} \right) \cos \left( \frac{3\theta - 5\theta}{2} \right) = 2 \cos(4\theta) \cos(-\theta) = 2 \cos(4\theta) \cos(\theta) \]

### Step 3: Combining the Results
Now substituting back into the original equation:
\[ 2 \cos(4\theta) \cos(3\theta) + 2 \cos(4\theta) \cos(\theta) = 0 \]
Factoring out \( 2 \cos(4\theta) \):
\[ 2 \cos(4\theta) (\cos(3\theta) + \cos(\theta)) = 0 \]

### Step 4: Setting Each Factor to Zero
We can set each factor to zero:
1. \( \cos(4\theta) = 0 \)
2. \( \cos(3\theta) + \cos(\theta) = 0 \)

#### Solving \( \cos(4\theta) = 0 \)
The general solution for \( \cos x = 0 \) is:
\[ 4\theta = \frac{(2n + 1)\pi}{2} \quad \Rightarrow \quad \theta = \frac{(2n + 1)\pi}{8} \]

#### Solving \( \cos(3\theta) + \cos(\theta) = 0 \)
This can be rewritten as:
\[ \cos(3\theta) = -\cos(\theta) \]
Using the identity \( \cos A = -\cos B \) gives:
\[ 3\theta = (2m + 1)\frac{\pi}{2} \quad \text{or} \quad 3\theta = 2m\pi \pm \theta \]
From \( 3\theta = (2m + 1)\frac{\pi}{2} \):
\[ \theta = \frac{(2m + 1)\pi}{6} \]

### Step 5: General Solution
The general values of \( \theta \) are:
1. From \( \cos(4\theta) = 0 \): \( \theta = \frac{(2n + 1)\pi}{8} \)
2. From \( \cos(3\theta) + \cos(\theta) = 0 \): \( \theta = \frac{(2m + 1)\pi}{6} \)

### Final Answer
Thus, the general values of \( \theta \) can be expressed as:
\[ \theta = \frac{(2n + 1)\pi}{8} \quad \text{and} \quad \theta = \frac{(2m + 1)\pi}{6} \]
where \( n \) and \( m \) are integers.