The general solution of `tan theta+tan 2 theta+tan 3 theta=0` is

Options

A. `theta = npi//6, n in Z` only
B. `theta=n pi pm alpha, n in Z`, where `tan alpha=1//sqrt(2)` only
C. Both a and b
D. none of these

Explanation

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

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

### Step 2: Use the tangent addition formula
We can express \( \tan 3\theta \) in terms of \( \tan \theta \) and \( \tan 2\theta \) using the tangent addition formula:
\[ \tan(3\theta) = \tan(\theta + 2\theta) = \frac{\tan \theta + \tan 2\theta}{1 - \tan \theta \tan 2\theta} \]
Substituting this into the equation gives us:
\[ \tan \theta + \frac{\tan \theta + \tan 2\theta}{1 - \tan \theta \tan 2\theta} = 0 \]

### Step 3: Combine terms
Let \( x = \tan \theta \) and \( y = \tan 2\theta \). We can rewrite the equation as:
\[ x + \frac{x + y}{1 - xy} = 0 \]
Multiplying through by \( (1 - xy) \) to eliminate the fraction:
\[ x(1 - xy) + x + y = 0 \]
This simplifies to:
\[ x - xy^2 + y = 0 \]

### Step 4: Factor the equation
Now we can rearrange the equation:
\[ x + y - xy^2 = 0 \]
This can be factored as:
\[ x + y = xy^2 \]

### Step 5: Solve for \( y \)
From the equation \( y = \tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} \), we can substitute \( y \) back into the equation. This gives us:
\[ \tan \theta + \frac{2\tan \theta}{1 - \tan^2 \theta} = 0 \]
Factoring out \( \tan \theta \):
\[ \tan \theta \left(1 + \frac{2}{1 - \tan^2 \theta}\right) = 0 \]

### Step 6: Set each factor to zero
This gives us two cases:
1. \( \tan \theta = 0 \)
2. \( 1 + \frac{2}{1 - \tan^2 \theta} = 0 \)

### Step 7: Solve for \( \theta \)
1. For \( \tan \theta = 0 \):
\[ \theta = n\pi, \quad n \in \mathbb{Z} \]

2. For \( 1 + \frac{2}{1 - \tan^2 \theta} = 0 \):
Rearranging gives:
\[ 1 - \tan^2 \theta + 2 = 0 \implies \tan^2 \theta = 3 \implies \tan \theta = \pm \sqrt{3} \]
This gives:
\[ \theta = n\pi + \frac{\pi}{3}, \quad n \in \mathbb{Z} \quad \text{and} \quad \theta = n\pi - \frac{\pi}{3}, \quad n \in \mathbb{Z} \]

### Final General Solution
Combining all solutions, we have:
\[ \theta = n\pi, \quad n \in \mathbb{Z} \]
\[ \theta = n\pi + \frac{\pi}{3}, \quad n \in \mathbb{Z} \]
\[ \theta = n\pi - \frac{\pi}{3}, \quad n \in \mathbb{Z} \]