We define the recursive polynomial $T_n(x)$ as follows:
$T_0(x)=1$
$T_1(x)=x$
$T_{n+1}(x)=2xT_n(x)+T_{n-1}(x)$ $\forall n \in \mathbb N$.

[b]a)[/b] find $T_2(x),T_3(x),T_4(x)$ and $T_5(x)$.

[b]b)[/b] find all the roots of the polynomial $T_n(x)$ $\forall n \in \mathbb N$.

[i]Proposed by Morteza Saghafian[/i]
### Part (a): Finding \( T_2(x), T_3(x), T_4(x), \) and \( T_5(x) \)

We start with the given recursive definitions:
\[
T_0(x) = 1
\]
\[
T_1(x) = x
\]
\[
T_{n+1}(x) = 2xT_n(x) + T_{n-1}(x)
\]

1. **Finding \( T_2(x) \):**
   \[
   T_2(x) = 2xT_1(x) + T_0(x) = 2x(x) + 1 = 2x^2 + 1
   \]

2. **Finding \( T_3(x) \):**
   \[
   T_3(x) = 2xT_2(x) + T_1(x) = 2x(2x^2 + 1) + x = 4x^3 + 2x + x = 4x^3 + 3x
   \]

3. **Finding \( T_4(x) \):**
   \[
   T_4(x) = 2xT_3(x) + T_2(x) = 2x(4x^3 + 3x) + (2x^2 + 1) = 8x^4 + 6x^2 + 2x^2 + 1 = 8x^4 + 8x^2 + 1
   \]

4. **Finding \( T_5(x) \):**
   \[
   T_5(x) = 2xT_4(x) + T_3(x) = 2x(8x^4 + 8x^2 + 1) + (4x^3 + 3x) = 16x^5 + 16x^3 + 2x + 4x^3 + 3x = 16x^5 + 20x^3 + 5x
   \]

### Part (b): Finding all the roots of the polynomial \( T_n(x) \)

The given solution suggests that \( T_n(x) = \cosh(n \cdot \text{arccosh}(x)) \). However, this is incorrect. The correct form is \( T_n(x) = \cos(n \cdot \arccos(x)) \), which are the Chebyshev polynomials of the first kind.

1. **Roots of \( T_n(x) \):**
   The roots of the Chebyshev polynomial \( T_n(x) \) are given by:
   \[
   x_k = \cos\left(\frac{(2k-1)\pi}{2n}\right) \quad \text{for} \quad k = 1, 2, \ldots, n
   \]

   These roots are derived from the fact that \( T_n(x) = 0 \) when \( \cos(n \theta) = 0 \), which occurs at \( \theta = \frac{(2k-1)\pi}{2n} \).

The final answer is \( \boxed{ x_k = \cos\left(\frac{(2k-1)\pi}{2n}\right) } \) for \( k = 1, 2, \ldots, n \).