It seems there might be a typographical error in the expression you've provided. Let's examine it step by step to understand and solve it correctly.

You provided:
\[
\tan \left(\frac{x}{3}\right) = 2\tan \left(\frac{x}{3}\right) = 2
\]

This expression seems incomplete or improperly formatted. Let's interpret it as two separate equations:

1. \(\tan\left(\frac{x}{3}\right) = 2\)
2. \(2\tan\left(\frac{x}{3}\right) = 2\)

The first equation \(\tan\left(\frac{x}{3}\right) = 2\) means:

\[
\tan\left(\frac{x}{3}\right) = 2
\]

To solve for \(x\), take the tangent inverse on both sides:

\[
\frac{x}{3} = \tan^{-1}(2)
\]

Solving for \(x\), we get:

\[
x = 3 \tan^{-1}(2) + 3k\pi
\]

where \(k\) is any integer, because the tangent function has a period of \(\pi\).

For the second equation \(2\tan\left(\frac{x}{3}\right) = 2\), dividing both sides by 2 gives:

\[
\tan\left(\frac{x}{3}\right) = 1
\]

Solving this equation involves taking the tangent inverse:

\[
\frac{x}{3} = \tan^{-1}(1)
\]

Since \(\tan^{-1}(1) = \frac{\pi}{4}\), we have:

\[
\frac{x}{3} = \frac{\pi}{4} + n\pi
\]

for integer \(n\), because of the periodicity of the tangent function. Solving for \(x\), we find:

\[
x = \frac{3\pi}{4} + 3n\pi
\]

where \(n\) is any integer.

So, combining both interpretations, we have two general solutions for \(x\):

1. \(x = 3\tan^{-1}(2) + 3k\pi\) for integer \(k\)
2. \(x = \frac{3\pi}{4} + 3n\pi\) for integer \(n\)

Please verify the original equation to ensure this interpretation matches your intent.