To solve the equation \(|x - 1| - 2x = \frac{1}{2}\), we need to consider the definition of the absolute value function, which splits the equation into two cases based on the value of \(x\).

### Case 1: \(x - 1 \geq 0\)
In this case, \(|x - 1| = x - 1\). The equation becomes:
\[ x - 1 - 2x = \frac{1}{2} \]

Simplify the equation:
\[ -x - 1 = \frac{1}{2} \]

Add 1 to both sides:
\[ -x = \frac{1}{2} + 1 \]
\[ -x = \frac{1}{2} + \frac{2}{2} \]
\[ -x = \frac{3}{2} \]

Multiply both sides by -1:
\[ x = -\frac{3}{2} \]

Since \(x - 1 \geq 0\) implies \(x \geq 1\), \(x = -\frac{3}{2}\) does not satisfy this condition. Therefore, there is no solution in this case.

### Case 2: \(x - 1 < 0\)
In this case, \(|x - 1| = -(x - 1) = -x + 1\). The equation becomes:
\[ -x + 1 - 2x = \frac{1}{2} \]

Simplify the equation:
\[ -3x + 1 = \frac{1}{2} \]

Subtract 1 from both sides:
\[ -3x = \frac{1}{2} - 1 \]
\[ -3x = \frac{1}{2} - \frac{2}{2} \]
\[ -3x = -\frac{1}{2} \]

Divide both sides by -3:
\[ x = \frac{-\frac{1}{2}}{-3} \]
\[ x = \frac{1}{2} \times \frac{1}{3} \]
\[ x = \frac{1}{6} \]

Since \(x - 1 < 0\) implies \(x < 1\), \(x = \frac{1}{6}\) satisfies this condition.

### Conclusion
The solution to the equation \(|x - 1| - 2x = \frac{1}{2}\) is:
\[ x = \frac{1}{6} \]