how to find the range for x for which the equation holds

Question

how to find the value of x for which the equation holds?

Answer 1

You have $\tan(\alpha) = x~$ and $~\cos(\beta) = \frac{1-x^2}{1+x^2}~$ with $2\alpha = \beta.$

Using the formula from my comment,

$$\tan(\beta) = \tan(2\alpha) = \frac{2x}{1-x^2} \implies \sin(\beta) = \tan(\beta)\times\cos(\beta) = \frac{2x}{1+x^2}.$$

Now use the pythagorean theorem to obtain a constraint on $x$.

$$\sin^2(\beta) + \cos^2(\beta) = 1 \implies \frac{4x^2}{1 + 2x^2 + x^4} + \frac{1 - 2x^2 + x^4}{1 + 2x^2 + x^4} = 1.$$

Clearly, this equation holds for any value of $x$.

Now contrast the question in your query's title with the question that you posed in your query. Presumably, the intended answer is that the equation holds for any value of $x$. Therefore, the allowable range of $x$ is any real number.

Addendum
Responding to the OP's latest comment.

First, consider $x = -1 \implies \alpha = (-\pi/4) \implies \beta = (-\pi/2) \implies 0 = \cos(\beta) = \frac{1 - [-1]^2}{1 + [-1]^2}.$

Therefore, the math clearly works.

However, if your book has adopted the arbitrary convention that the domain of the Arccosine function is $[0,\pi]$, then the sine of $\beta$ is not permitted to be negative.

Therefore, since $~\sin(\beta) = \frac{2x}{1+x^2},~$the only way to satisfy the arbitrary convention is to insist that $x$ be non-negative.