The equation you've written appears to have a typographical error. It seems you meant to write:

\[ 2\sin(3x + 1) = 1 \]

Let's solve this equation step by step:

1. **Divide both sides by 2:**
\[
\sin(3x + 1) = \frac{1}{2}
\]

2. **Identify the general solutions for \(\sin\) function:**
The sine function is equal to \( \frac{1}{2} \) at the angles:
\[
3x + 1 = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad 3x + 1 = \frac{5\pi}{6} + 2k\pi
\]
where \( k \) is any integer.

3. **Solve for \( x \):**

For the first case:
\[
3x + 1 = \frac{\pi}{6} + 2k\pi
\]
Subtract 1 from both sides:
\[
3x = \frac{\pi}{6} - 1 + 2k\pi
\]
Divide by 3:
\[
x = \frac{1}{3}\left(\frac{\pi}{6} - 1 + 2k\pi\right) = \frac{\pi}{18} - \frac{1}{3} + \frac{2k\pi}{3}
\]

For the second case:
\[
3x + 1 = \frac{5\pi}{6} + 2k\pi
\]
Subtract 1 from both sides:
\[
3x = \frac{5\pi}{6} - 1 + 2k\pi
\]
Divide by 3:
\[
x = \frac{1}{3}\left(\frac{5\pi}{6} - 1 + 2k\pi\right) = \frac{5\pi}{18} - \frac{1}{3} + \frac{2k\pi}{3}
\]

4. **Combine the solutions:**
The complete solution for the equation is:
\[
x = \frac{\pi}{18} - \frac{1}{3} + \frac{2k\pi}{3}, \quad x = \frac{5\pi}{18} - \frac{1}{3} + \frac{2k\pi}{3} \quad (k \in \mathbb{Z})
\]

These expressions represent all solutions for \( x \) corresponding to the given trigonometric equation \( 2\sin(3x + 1) = 1 \).