Find all real numbers $a,b,c,d,e$ in the interval $[-2,2]$, that satisfy:
\begin{align*}a+b+c+d+e &= 0\\ a^3+b^3+c^3+d^3+e^3&= 0\\ a^5+b^5+c^5+d^5+e^5&=10 \end{align*}
To find all real numbers \(a, b, c, d, e\) in the interval \([-2, 2]\) that satisfy the given equations:
\[
\begin{align*}
a + b + c + d + e &= 0, \\
a^3 + b^3 + c^3 + d^3 + e^3 &= 0, \\
a^5 + b^5 + c^5 + d^5 + e^5 &= 10,
\end{align*}
\]
we will proceed as follows:

1. **Substitution using trigonometric identities:**
   Let \(a = 2\cos\alpha\), \(b = 2\cos\beta\), \(c = 2\cos\gamma\), \(d = 2\cos\delta\), and \(e = 2\cos\epsilon\). This substitution ensures that \(a, b, c, d, e \in [-2, 2]\).

2. **Sum of cosines:**
   Given \(a + b + c + d + e = 0\), we have:
   \[
   2(\cos\alpha + \cos\beta + \cos\gamma + \cos\delta + \cos\epsilon) = 0 \implies \cos\alpha + \cos\beta + \cos\gamma + \cos\delta + \cos\epsilon = 0.
   \]

3. **Sum of cubes:**
   Using the identity for the sum of cubes:
   \[
   a^3 + b^3 + c^3 + d^3 + e^3 = 0,
   \]
   we substitute \(a = 2\cos\alpha\), etc., and use the triple angle formula:
   \[
   (2\cos\alpha)^3 + (2\cos\beta)^3 + (2\cos\gamma)^3 + (2\cos\delta)^3 + (2\cos\epsilon)^3 = 0.
   \]
   Simplifying, we get:
   \[
   8(\cos^3\alpha + \cos^3\beta + \cos^3\gamma + \cos^3\delta + \cos^3\epsilon) = 0 \implies \cos^3\alpha + \cos^3\beta + \cos^3\gamma + \cos^3\delta + \cos^3\epsilon = 0.
   \]

4. **Sum of fifth powers:**
   Given \(a^5 + b^5 + c^5 + d^5 + e^5 = 10\), we substitute \(a = 2\cos\alpha\), etc., and use the quintuple angle formula:
   \[
   (2\cos\alpha)^5 + (2\cos\beta)^5 + (2\cos\gamma)^5 + (2\cos\delta)^5 + (2\cos\epsilon)^5 = 10.
   \]
   Simplifying, we get:
   \[
   32(\cos^5\alpha + \cos^5\beta + \cos^5\gamma + \cos^5\delta + \cos^5\epsilon) = 10 \implies \cos^5\alpha + \cos^5\beta + \cos^5\gamma + \cos^5\delta + \cos^5\epsilon = \frac{10}{32} = \frac{5}{16}.
   \]

5. **Solving the polynomial:**
   We need to find the roots of the polynomial \(x^5 - 5x^3 + 5x - 2 = 0\). The roots of this polynomial are:
   \[
   x = 2, \quad x = \frac{\sqrt{5} - 1}{2}, \quad x = -\frac{1 + \sqrt{5}}{2}.
   \]
   These roots must satisfy the given conditions. By the irrationality of \(\sqrt{5}\) and the condition \(a + b + c + d + e = 0\), the roots must come in pairs.

6. **Possible values:**
   The possible values for \(a, b, c, d, e\) are:
   \[
   (a, b, c, d, e) = \left(2, \frac{\sqrt{5} - 1}{2}, \frac{\sqrt{5} - 1}{2}, -\frac{1 + \sqrt{5}}{2}, -\frac{1 + \sqrt{5}}{2}\right).
   \]
   This indeed satisfies the given equations.

The final answer is \( \boxed{ \left(2, \frac{\sqrt{5} - 1}{2}, \frac{\sqrt{5} - 1}{2}, -\frac{1 + \sqrt{5}}{2}, -\frac{1 + \sqrt{5}}{2}\right) } \).