To solve the equation
\[ \frac{1}{x} = \frac{x + 3}{2x^2}, \]
we can start by cross-multiplying to eliminate the fractions:
\[ 1 \cdot 2x^2 = (x + 3) \cdot x. \]
This gives us:
\[ 2x^2 = x^2 + 3x. \]
Next, we can rearrange the equation by moving all terms to one side:
\[ 2x^2 - x^2 - 3x = 0, \]
which simplifies to:
\[ x^2 - 3x = 0. \]
Now, we can factor out an \( x \):
\[ x(x - 3) = 0. \]
Setting each factor to zero gives us the possible solutions:
- \( x = 0 \)
- \( x - 3 = 0 \) which leads to \( x = 3 \).
Thus, the solutions to the equation are \( x = 0 \) and \( x = 3 \).
Now, let's check if these solutions are valid in the original equation:
- For \( x = 0 \), \(\frac{1}{0}\) is undefined, so \( x = 0\) is not a valid solution.
- For \( x = 3 \), substituting into the original equation we have:
\[ \frac{1}{3} = \frac{3 + 3}{2 \cdot 3^2} = \frac{6}{18} = \frac{1}{3}, \]
which is valid.
Thus, the only valid solution is:
\[ x = 3. \]
The correct answer is: \( x = 3 \).