According to an extension of Niven's theorem , the only rational values of tangent are $0$ and $\pm 1$ . Since the equation can be rearranged to $x = \tan (\tan (x))$ (and if we assume $x$ must be an integer), we only need to test these three values, and we find:

$\tan(-1) \neq \tan^{-1}(-1)$

$\tan(0) = \tan^{-1}(0)$

$\tan(1) \neq \tan^{-1}(1)$

Therefore, $\boxed{x = 0}$ .

$\tan(x) = \tan^{-1}(x)$ $x = \tan(\tan(x))$

According to an extension of Niven's Theorem, the only rational values of tangent are $0$ and $1 \pm 1$ .

When graphed, the only line that exists is the one with $x = 0$ (As seen in the pictures below)

$\tan(0) = \tan^{-1}(0)$ $\tan(1) \cancel{=} \tan^{-1}(1)$ $\tan(-1) \cancel{=} \tan^{-1}(-1)$

$x=0$

At first glance I saw ‘positive’, and I was like WHAT? So I tried re-writing the formula as $\tan (\tan (x))=x$ , etc. One solution was $x=0$ . Then I saw that the wording has changed, and realised the only possibility was that the answer 0 isn’t considered positive and the wording had to be changed, so my proof is $\tan (x)=\tan ^{-1} (x)\Rightarrow \tan ( \tan (x))=x,$ $\text{and the only integer solution to this equation is 0 since 0 is a valid solution and the solution to this equation has an irrational period.}$ Since the period is irrational, no matter what integer you multiply it by, the result is irrational.

$\tan(0) = 0$ , $\tan^{-1}(0)=0$ therefore the answer is $\boxed{0}$ . It is debatable whether 0 is positive but in this case it is.

Edit: the wording of the question has been changed so 0 being positive or not can be left aside