The problem consists of two parts. First, express the function $f(x) = \cos(2x) + \sqrt{3}\sin(2x)$ in the form $f(x) = \lambda \cos(ax + b)$, and determine the values of $a$, $b$, and $\lambda$. Second, solve the inequality $f(x) \ge \sqrt{3}$ in the interval $[0, \pi]$.

1. Problem Description

The problem consists of two parts.

First, express the function $f(x) = \cos(2x) + \sqrt{3}\sin(2x)$ in the form $f(x) = \lambda \cos(ax + b)$, and determine the values of $a$, $b$, and $\lambda$.

Second, solve the inequality $f(x) \ge \sqrt{3}$ in the interval $[0, \pi]$.

2. Solution Steps

Part 1: Express $f(x)$ in the form $\lambda \cos(ax + b)$.

We have $f(x) = \cos(2x) + \sqrt{3}\sin(2x)$. We want to express this as $\lambda \cos(ax + b)$.

We can rewrite $\lambda \cos(ax + b)$ as $\lambda (\cos(ax) \cos(b) - \sin(ax) \sin(b)) = \lambda \cos(b) \cos(ax) - \lambda \sin(b) \sin(ax)$.

Comparing the two expressions, we have:

$\cos(2x) + \sqrt{3}\sin(2x) = \lambda \cos(b) \cos(ax) - \lambda \sin(b) \sin(ax)$.

From this, we can infer that $a = 2$.

Then we have:

$\cos(2x) + \sqrt{3}\sin(2x) = \lambda \cos(b) \cos(2x) - \lambda \sin(b) \sin(2x)$.

Equating the coefficients of $\cos(2x)$ and $\sin(2x)$, we get:

$\lambda \cos(b) = 1$ and $-\lambda \sin(b) = \sqrt{3}$.

Squaring and adding the two equations, we have:

$(\lambda \cos(b))^2 + (-\lambda \sin(b))^2 = 1^2 + (\sqrt{3})^2$.

$\lambda^2 (\cos^2(b) + \sin^2(b)) = 1 + 3 = 4$.

Since $\cos^2(b) + \sin^2(b) = 1$, we have $\lambda^2 = 4$.

Since $\lambda > 0$, we take $\lambda = 2$.

Now we have $2 \cos(b) = 1$ and $-2 \sin(b) = \sqrt{3}$.

Thus $\cos(b) = \frac{1}{2}$ and $\sin(b) = -\frac{\sqrt{3}}{2}$.

This means that $b = -\frac{\pi}{3}$.

Therefore, $f(x) = 2 \cos(2x - \frac{\pi}{3})$.

Part 2: Solve the inequality $f(x) \ge \sqrt{3}$ in the interval $[0, \pi]$.

We have $2 \cos(2x - \frac{\pi}{3}) \ge \sqrt{3}$.

$\cos(2x - \frac{\pi}{3}) \ge \frac{\sqrt{3}}{2}$.

Let $u = 2x - \frac{\pi}{3}$.

We want to solve $\cos(u) \ge \frac{\sqrt{3}}{2}$.

We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$.

The general solution is $-\frac{\pi}{6} + 2k\pi \le u \le \frac{\pi}{6} + 2k\pi$ for some integer $k$.

So $-\frac{\pi}{6} + 2k\pi \le 2x - \frac{\pi}{3} \le \frac{\pi}{6} + 2k\pi$.

Adding $\frac{\pi}{3}$ to all parts of the inequality, we have:

$\frac{\pi}{6} + 2k\pi \le 2x \le \frac{\pi}{2} + 2k\pi$.

Dividing by 2, we have:

$\frac{\pi}{12} + k\pi \le x \le \frac{\pi}{4} + k\pi$.

We want to find the values of $x$ in the interval $[0, \pi]$.

If $k = 0$, then $\frac{\pi}{12} \le x \le \frac{\pi}{4}$.

If $k = 1$, then $\frac{\pi}{12} + \pi \le x \le \frac{\pi}{4} + \pi$, which is $\frac{13\pi}{12} \le x \le \frac{5\pi}{4}$.

Since we are only interested in solutions in the interval $[0, \pi]$, we have $\frac{\pi}{12} \le x \le \frac{\pi}{4}$.

3. Final Answer

$f(x) = 2\cos(2x - \frac{\pi}{3})$ where $a = 2, b = -\frac{\pi}{3}, \lambda = 2$.

The solution to $f(x) \ge \sqrt{3}$ in $[0, \pi]$ is $[\frac{\pi}{12}, \frac{\pi}{4}]$.