To solve the equation $\cos(2x) = \frac{1}{2}$ in the interval $[0, 2\pi)$, we need to find all $x$ such that the given equation holds within this interval. Here are the steps to solve this equation:

Step 1: Solve for $2x$

We know: $\cos(2x) = \frac{1}{2}$ The general solution for $\cos(\theta) = \frac{1}{2}$ is: $\theta = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad \theta = \frac{5\pi}{3} + 2n\pi$ where $n$ is any integer.

Step 2: Substitute $2x = \theta$

Since $2x = \theta$, we have two equations:

1. $2x = \frac{\pi}{3} + 2n\pi$
2. $2x = \frac{5\pi}{3} + 2n\pi$

Step 3: Solve for $x$ in both cases

For each case, divide by 2 to isolate $x$.

Case 1

$x = \frac{\pi}{6} + n\pi$

Case 2

$x = \frac{5\pi}{6} + n\pi$

Step 4: Find values of $x$ in the interval $[0, 2\pi)$

For $x = \frac{\pi}{6} + n\pi$:

1. When $n = 0$: $x = \frac{\pi}{6}$
2. When $n = 1$: $x = \frac{\pi}{6} + \pi = \frac{7\pi}{6}$

For $x = \frac{5\pi}{6} + n\pi$:

1. When $n = 0$: $x = \frac{5\pi}{6}$
2. When $n = 1$: $x = \frac{5\pi}{6} + \pi = \frac{11\pi}{6}$

Step 5: List all solutions

The solutions for $x$ in the interval $[0, 2\pi)$ are: $x = \frac{\pi}{6}, \quad \frac{7\pi}{6}, \quad \frac{5\pi}{6}, \quad \frac{11\pi}{6}$

These are all the values of $x$ that satisfy $\cos(2x) = \frac{1}{2}$ in the interval $[0, 2\pi)$.

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Would you like further explanation on any steps? Here are five related questions:

1. How would the solutions change if the interval was $[0, 4\pi)$?
2. What is the process for solving similar equations involving $\sin(2x) = \frac{1}{2}$?
3. How does the double-angle formula apply in such trigonometric equations?
4. What are the general steps to find solutions for trigonometric equations involving transformations?
5. How can we verify the solutions we found are correct?

Tip: Always remember to check the interval specified, as trigonometric equations often have multiple valid solutions across different intervals.