We are tasked with solving the equation:

\[
2\cos(4x) - 1 = 0
\]

### Step 1: Isolate the cosine function
First, add 1 to both sides:

\[
2\cos(4x) = 1
\]

Next, divide both sides by 2:

\[
\cos(4x) = \frac{1}{2}
\]

### Step 2: Solve for \(4x\)
We now need to find the angles \(4x\) for which \(\cos(4x) = \frac{1}{2}\). The cosine function equals \(\frac{1}{2}\) at angles \( \theta = \frac{\pi}{3} \) and \( \theta = -\frac{\pi}{3} \), and all coterminal angles (which are multiples of \(2\pi\)).

Thus, we have the general solutions:

\[
4x = 2n\pi \pm \frac{\pi}{3} \quad \text{for any integer } n.
\]

### Step 3: Solve for \(x\)
To isolate \(x\), divide by 4:

\[
x = \frac{2n\pi \pm \frac{\pi}{3}}{4}
\]

Simplifying:

\[
x = \frac{n\pi}{2} \pm \frac{\pi}{12}
\]

This gives the general solution:

\[
x = \frac{n\pi}{2} + \frac{\pi}{12} \quad \text{or} \quad x = \frac{n\pi}{2} - \frac{\pi}{12}, \quad n \in \mathbb{Z}.
\]

### Final Answer:
The solutions are:

\[
x = \frac{n\pi}{2} \pm \frac{\pi}{12}, \quad n \in \mathbb{Z}.
\]

Would you like further details on any step or have additional questions?

### Related Questions:
1. How do we solve equations with different trigonometric identities?
2. What are the solutions to \(\cos(2x) = 0\)?
3. Can cosine equations have no real solutions?
4. What is the importance of periodicity in trigonometric solutions?
5. How do phase shifts affect the solutions of trigonometric equations?

### Tip:
Always remember the periodic nature of trigonometric functions when solving equations, which leads to multiple solutions!

Learn how to solve the trigonometric equation 2cos(4x) - 1 = 0 step by step. This solution covers key concepts in trigonometry, including cosine functions and periodicity. Ideal for students in Grades 10-12.

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