If `tanx+tan((pi)/(3)-x)+tan((2pi)/(3)+x)=3`, then

Options

A. tan x =1
B. tan 2x =1
C. tan 3x = 1
D. None of these

Explanation

To solve the equation \( \tan x + \tan\left(\frac{\pi}{3} - x\right) + \tan\left(\frac{2\pi}{3} + x\right) = 3 \), we will use the properties of tangent and the tangent addition formula.

### Step 1: Rewrite the terms using tangent subtraction and addition formulas
We know that:
\[ \tan\left(\frac{\pi}{3} - x\right) = \frac{\tan\frac{\pi}{3} - \tan x}{1 + \tan\frac{\pi}{3} \tan x} = \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} \]
and
\[ \tan\left(\frac{2\pi}{3} + x\right) = \tan\left(\pi - \frac{\pi}{3} + x\right) = -\tan\left(\frac{\pi}{3} - x\right) = -\frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} \]

### Step 2: Substitute these into the original equation
Substituting these into the original equation gives:
\[ \tan x + \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} - \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} = 3 \]
Notice that the terms involving \(\tan\left(\frac{2\pi}{3} + x\right)\) cancel out, simplifying our equation to:
\[ \tan x = 3 \]

### Step 3: Solve for \(x\)
Now, we can solve for \(x\):
\[ \tan x = 3 \]
This implies:
\[ x = \tan^{-1}(3) + n\pi, \quad n \in \mathbb{Z} \]

### Final Answer
Thus, the general solution for \(x\) is:
\[ x = \tan^{-1}(3) + n\pi, \quad n \in \mathbb{Z} \]