Solve `cos theta+cos 2theta+cos 3theta=0`.

Explanation

To solve the equation \( \cos \theta + \cos 2\theta + \cos 3\theta = 0 \), we will use trigonometric identities and algebraic manipulation. Here’s a step-by-step solution:

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

### Step 2: Group terms
We can group the terms as follows:
\[ \cos \theta + \cos 3\theta + \cos 2\theta = 0 \]

### Step 3: Use the cosine addition formula
Using the identity for the sum of cosines, we have:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Let \( A = \theta \) and \( B = 3\theta \):
\[ \cos \theta + \cos 3\theta = 2 \cos\left(\frac{\theta + 3\theta}{2}\right) \cos\left(\frac{\theta - 3\theta}{2}\right) = 2 \cos(2\theta) \cos(\theta) \]
Thus, we can rewrite the equation as:
\[ 2 \cos(2\theta) \cos(\theta) + \cos 2\theta = 0 \]

### Step 4: Factor out \(\cos 2\theta\)
Now, factor out \(\cos 2\theta\):
\[ \cos 2\theta (2 \cos \theta + 1) = 0 \]

### Step 5: Set each factor to zero
Now we can set each factor to zero:
1. \( \cos 2\theta = 0 \)
2. \( 2 \cos \theta + 1 = 0 \)

### Step 6: Solve \( \cos 2\theta = 0 \)
The general solution for \( \cos x = 0 \) is:
\[ x = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus, for \( 2\theta \):
\[ 2\theta = \frac{\pi}{2} + n\pi \implies \theta = \frac{\pi}{4} + \frac{n\pi}{2} \]

### Step 7: Solve \( 2 \cos \theta + 1 = 0 \)
From \( 2 \cos \theta + 1 = 0 \):
\[ 2 \cos \theta = -1 \implies \cos \theta = -\frac{1}{2} \]
The general solution for \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad \theta = \frac{4\pi}{3} + 2n\pi \]

### Step 8: Combine solutions
Thus, the complete solution set is:
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 solutions to the equation \( \cos \theta + \cos 2\theta + \cos 3\theta = 0 \) are:
\[ \theta = \frac{\pi}{4} + \frac{n\pi}{2}, \quad \theta = \frac{2\pi}{3} + 2n\pi, \quad \theta = \frac{4\pi}{3} + 2n\pi \quad (n \in \mathbb{Z}) \]