To solve the equation

\[
|2x + \frac{1}{2}| + |2x - \frac{3}{2}| = 2
\]

we start by determining the critical points where the expressions inside the absolute values change sign. This occurs when \(2x + \frac{1}{2} = 0\) and \(2x - \frac{3}{2} = 0\).

1. Solving \(2x + \frac{1}{2} = 0\):

\[
2x = -\frac{1}{2} \implies x = -\frac{1}{4}
\]

2. Solving \(2x - \frac{3}{2} = 0\):

\[
2x = \frac{3}{2} \implies x = \frac{3}{4}
\]

These critical points split the number line into the following intervals:

- \( (-\infty, -\frac{1}{4}) \)
- \( [-\frac{1}{4}, \frac{3}{4}) \)
- \( [\frac{3}{4}, \infty) \)

Next, we will evaluate the absolute value expressions in each interval.

### Interval 1: \(x < -\frac{1}{4}\)

In this interval, both \(2x + \frac{1}{2}\) and \(2x - \frac{3}{2}\) are negative, so we have:

\[
|2x + \frac{1}{2}| = -(2x + \frac{1}{2}) = -2x - \frac{1}{2}
\]
\[
|2x - \frac{3}{2}| = -(2x - \frac{3}{2}) = -2x + \frac{3}{2}
\]

Substituting these into the equation gives:

\[
(-2x - \frac{1}{2}) + (-2x + \frac{3}{2}) = 2
\]

This simplifies to:

\[
-4x + 1 = 2 \implies -4x = 1 \implies x = -\frac{1}{4}
\]

However, this value does not fall within the interval \( (-\infty, -\frac{1}{4})\).

### Interval 2: \(-\frac{1}{4} \leq x < \frac{3}{4}\)

In this interval, \(2x + \frac{1}{2} \geq 0\) and \(2x - \frac{3}{2} < 0\), leading to:

\[
|2x + \frac{1}{2}| = 2x + \frac{1}{2}
\]
\[
|2x - \frac{3}{2}| = -(2x - \frac{3}{2}) = -2x + \frac{3}{2}
\]

Thus, the equation becomes:

\[
(2x + \frac{1}{2}) + (-2x + \frac{3}{2}) = 2
\]

This simplifies to:

\[
2 = 2
\]

This is true for any \(x\) in the interval \( \left[-\frac{1}{4}, \frac{3}{4}\right) \).

### Interval 3: \(x \geq \frac{3}{4}\)

In this interval, both expressions are non-negative:

\[
|2x + \frac{1}{2}| = 2x + \frac{1}{2}
\]
\[
|2x - \frac{3}{2}| = 2x - \frac{3}{2}
\]

So the equation becomes:

\[
(2x + \frac{1}{2}) + (2x - \frac{3}{2}) = 2
\]

This simplifies to:

\[
4x - 1 = 2 \implies 4x = 3 \implies x = \frac{3}{4}
\]

Since \(x = \frac{3}{4}\) is on the boundary of the interval, we accept it as a solution.

### Final Solution

Therefore, the solutions to the equation are:

\[
\boxed{[-\frac{1}{4}, \frac{3}{4}]}
\]