Let's solve the given problem step-by-step.

1. Problem Breakdown:

The function is $f(x) = 2\cos(x)$. We are asked:

1. To find the Lagrange interpolation polynomial $P_2(x)$ of degree 2 that interpolates $f(x)$ at points $0$, $\pi$, and $2\pi$.
2. To compute $P_2\left(\frac{3\pi}{4}\right)$.
3. To find an upper bound for the interpolation error.

Step 1: Lagrange Interpolation Polynomial

The general formula for the Lagrange interpolation polynomial is: $P_2(x) = \sum_{i=0}^2 f(x_i) \ell_i(x),$ where $\ell_i(x)$ is the Lagrange basis polynomial: $\ell_i(x) = \prod_{\substack{j=0 \\ j \neq i}}^2 \frac{x - x_j}{x_i - x_j}.$ Here the interpolation points are: $x_0 = 0, \quad x_1 = \pi, \quad x_2 = 2\pi.$

Step 1.1: Compute $f(x_i)$

$f(0) = 2\cos(0) = 2, \quad f(\pi) = 2\cos(\pi) = -2, \quad f(2\pi) = 2\cos(2\pi) = 2.$

Step 1.2: Basis Polynomials $\ell_i(x)$

• For $\ell_0(x)$: $\ell_0(x) = \frac{(x - \pi)(x - 2\pi)}{(0 - \pi)(0 - 2\pi)} = \frac{(x - \pi)(x - 2\pi)}{\pi(2\pi)} = \frac{(x - \pi)(x - 2\pi)}{2\pi^2}.$

• For $\ell_1(x)$: [ \ell_1(x) = \frac{(x - 0)(x - 2\pi)}