The problem asks us to solve the trigonometric equation:

$2 \cos^2(x) - 5 \cos(x) + 2 = 0, \quad \text{for } 0 \leq x < 2\pi.$

Step-by-step solution:

1. Rewrite the equation:
   Let $y = \cos(x)$. The equation becomes: $2y^2 - 5y + 2 = 0.$

2. Solve the quadratic equation:
   Using the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 2$, $b = -5$, and $c = 2$.

   Substitute these values: $y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} = \frac{5 \pm \sqrt{25 - 16}}{4}.$ $y = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4}.$

   This gives two solutions for $y$: $y = \frac{5 + 3}{4} = 2 \quad \text{(not valid, since } -1 \leq \cos(x) \leq 1\text{)},$ and $y = \frac{5 - 3}{4} = \frac{2}{4} = 0.5.$

   So, $\cos(x) = 0.5$.

3. Solve for $x$: Recall the general solution for $\cos(x) = 0.5$ in the interval $0 \leq x < 2\pi$: $x = \cos^{-1}(0.5).$ The solutions are: $x = \frac{\pi}{3}, \quad x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}.$

4. Write the final solution: The solutions to the equation are: $x = \frac{\pi}{3}, \frac{5\pi}{3}.$

   In decimal form (to two decimal places): $x \approx 1.05, 5.24.$

Final Answer:

$x = 1.05, 5.24$

Would you like a detailed explanation of any step or have further questions?

Here are five related questions to deepen your understanding:

1. What is the domain and range of $\cos(x)$?
2. How can you verify that $\cos(x) = 2$ is invalid?
3. How does the quadratic formula work in trigonometric contexts?
4. What are other methods to solve quadratic trigonometric equations?
5. Can you graph $y = 2\cos^2(x) - 5\cos(x) + 2$ to visualize the solution?

Tip: Always check the interval restrictions when solving trigonometric equations!