Solve tan `x+tan 2x + sqrt3 tan x tan 2x= sqrt3`

Answer

To solve the equation \( \tan x + \tan 2x + \sqrt{3} \tan x \tan 2x = \sqrt{3} \), we can follow these steps:

### Step 1: Rearrange the equation
Start by rewriting the equation to isolate the terms involving tangent:
\[ \tan x + \tan 2x = \sqrt{3} - \sqrt{3} \tan x \tan 2x \]
This can be simplified to:
\[ \tan x + \tan 2x = \sqrt{3}(1 - \tan x \tan 2x) \]

### Step 2: Use the tangent addition formula
Recall the tangent addition formula:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
In our case, let \( a = x \) and \( b = 2x \). Thus, we can express the left-hand side as:
\[ \tan(x + 2x) = \tan(3x) \]
So, we have:
\[ \tan(3x) = \sqrt{3} \]

### Step 3: Solve for \( 3x \)
We know that \( \tan \theta = \sqrt{3} \) when \( \theta = \frac{\pi}{3} + n\pi \) for any integer \( n \). Therefore, we can write:
\[ 3x = \frac{\pi}{3} + n\pi \]

### Step 4: Solve for \( x \)
Now, divide both sides by 3 to find \( x \):
\[ x = \frac{\pi}{9} + \frac{n\pi}{3} \]

### Final Solution
Thus, the general solution for \( x \) is:
\[ x = n\frac{\pi}{3} + \frac{\pi}{9} \]
where \( n \) is any integer.

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