Solve the general vlaue.
`2 cos theta + cos 3 theta =0`

Answer

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

### Step 1: Rewrite \(\cos 3\theta\)
We know that the cosine of triple angle can be expressed as:
\[ \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \]
Substituting this into the original equation gives:
\[ 2 \cos \theta + (4 \cos^3 \theta - 3 \cos \theta) = 0 \]

### Step 2: Simplify the equation
Now, we can simplify the equation:
\[ 2 \cos \theta + 4 \cos^3 \theta - 3 \cos \theta = 0 \]
This simplifies to:
\[ 4 \cos^3 \theta - \cos \theta = 0 \]

### Step 3: Factor out \(\cos \theta\)
We can factor out \(\cos \theta\):
\[ \cos \theta (4 \cos^2 \theta - 1) = 0 \]

### Step 4: Set each factor to zero
Now, we have two cases to solve:
1. \(\cos \theta = 0\)
2. \(4 \cos^2 \theta - 1 = 0\)

### Step 5: Solve \(\cos \theta = 0\)
For the first case:
\[ \cos \theta = 0 \]
This occurs at:
\[ \theta = \frac{(2n + 1) \pi}{2}, \quad n \in \mathbb{Z} \]

### Step 6: Solve \(4 \cos^2 \theta - 1 = 0\)
For the second case, we can rearrange it:
\[ 4 \cos^2 \theta = 1 \]
Dividing both sides by 4 gives:
\[ \cos^2 \theta = \frac{1}{4} \]
Taking the square root of both sides gives:
\[ \cos \theta = \frac{1}{2} \quad \text{or} \quad \cos \theta = -\frac{1}{2} \]

### Step 7: Find angles for \(\cos \theta = \frac{1}{2}\)
The solutions for \(\cos \theta = \frac{1}{2}\) are:
\[ \theta = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad \theta = -\frac{\pi}{3} + 2n\pi, \quad n \in \mathbb{Z} \]

### Step 8: Find angles for \(\cos \theta = -\frac{1}{2}\)
The solutions for \(\cos \theta = -\frac{1}{2}\) are:
\[ \theta = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad \theta = -\frac{2\pi}{3} + 2n\pi, \quad n \in \mathbb{Z} \]

### Final Solution
Combining all the solutions, we have:
\[ \theta = \frac{(2n + 1) \pi}{2} \quad \text{or} \quad \theta = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad \theta = -\frac{\pi}{3} + 2n\pi \quad \text{or} \quad \theta = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad \theta = -\frac{2\pi}{3} + 2n\pi, \quad n \in \mathbb{Z} \]