There exist two positive numbers $ x$ such that $ \sin(\arccos(\tan(\arcsin x)))\equal{}x$. Find the product of the two possible $ x$.
1. Let $\arcsin x = \theta$. Then, $x = \sin \theta$ and $\theta \in [0, \frac{\pi}{2}]$ since $x$ is a positive number.
2. We need to find $\tan(\arcsin x)$. Using the right triangle representation, we have:
   \[
   \sin \theta = x \quad \text{and} \quad \cos \theta = \sqrt{1 - x^2}
   \]
   Therefore,
   \[
   \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{x}{\sqrt{1 - x^2}}
   \]
3. Let $\beta = \arccos \left( \frac{x}{\sqrt{1 - x^2}} \right)$. Then, $\cos \beta = \frac{x}{\sqrt{1 - x^2}}$.
4. We need to find $\sin \beta$. Using the right triangle representation again, we have:
   \[
   \cos \beta = \frac{x}{\sqrt{1 - x^2}} \quad \text{and} \quad \sin \beta = \sqrt{1 - \cos^2 \beta}
   \]
   Therefore,
   \[
   \sin \beta = \sqrt{1 - \left( \frac{x}{\sqrt{1 - x^2}} \right)^2} = \sqrt{1 - \frac{x^2}{1 - x^2}} = \sqrt{\frac{1 - x^2 - x^2}{1 - x^2}} = \sqrt{\frac{1 - 2x^2}{1 - x^2}}
   \]
5. Given that $\sin(\arccos(\tan(\arcsin x))) = x$, we have:
   \[
   \sqrt{\frac{1 - 2x^2}{1 - x^2}} = x
   \]
6. Squaring both sides, we get:
   \[
   \left( \sqrt{\frac{1 - 2x^2}{1 - x^2}} \right)^2 = x^2 \implies \frac{1 - 2x^2}{1 - x^2} = x^2
   \]
7. Multiplying both sides by $1 - x^2$, we obtain:
   \[
   1 - 2x^2 = x^2 (1 - x^2) \implies 1 - 2x^2 = x^2 - x^4
   \]
8. Rearranging terms, we get:
   \[
   x^4 - 3x^2 + 1 = 0
   \]
9. Let $y = x^2$. Then, the equation becomes:
   \[
   y^2 - 3y + 1 = 0
   \]
10. Solving this quadratic equation using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get:
    \[
    y = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}
    \]
11. Therefore, the solutions for $x^2$ are:
    \[
    x^2 = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad x^2 = \frac{3 - \sqrt{5}}{2}
    \]
12. Since $x$ is positive, we take the positive square roots:
    \[
    x = \sqrt{\frac{3 + \sqrt{5}}{2}} \quad \text{and} \quad x = \sqrt{\frac{3 - \sqrt{5}}{2}}
    \]
13. We are asked to find the product of the two possible values of $x$. Therefore, we calculate:
    \[
    \left( \sqrt{\frac{3 + \sqrt{5}}{2}} \right) \left( \sqrt{\frac{3 - \sqrt{5}}{2}} \right) = \sqrt{\left( \frac{3 + \sqrt{5}}{2} \right) \left( \frac{3 - \sqrt{5}}{2} \right)}
    \]
14. Simplifying the product inside the square root:
    \[
    \left( \frac{3 + \sqrt{5}}{2} \right) \left( \frac{3 - \sqrt{5}}{2} \right) = \frac{(3 + \sqrt{5})(3 - \sqrt{5})}{4} = \frac{9 - 5}{4} = \frac{4}{4} = 1
    \]
15. Therefore, the product of the two possible values of $x$ is:
    \[
    \sqrt{1} = 1
    \]

The final answer is $\boxed{1}$