If $$A = \begin{bmatrix} 0 & \sin\alpha \\ \sin\alpha & 0 \end{bmatrix}$$ and $$\det\left(A^2 - \frac{1}{2}I\right) = 0$$, then a possible value of $$\alpha$$ is:

We have $$A = \begin{bmatrix} 0 & \sin\alpha \\ \sin\alpha & 0 \end{bmatrix}$$.

First, compute $$A^2$$: $$A^2 = \begin{bmatrix} 0 & \sin\alpha \\ \sin\alpha & 0 \end{bmatrix} \begin{bmatrix} 0 & \sin\alpha \\ \sin\alpha & 0 \end{bmatrix} = \begin{bmatrix} \sin^2\alpha & 0 \\ 0 & \sin^2\alpha \end{bmatrix} = \sin^2\alpha \cdot I$$.

Now compute $$A^2 - \frac{1}{2}I = \sin^2\alpha \cdot I - \frac{1}{2}I = \left(\sin^2\alpha - \frac{1}{2}\right)I$$.

The determinant of a scalar multiple of the identity matrix: $$\det(kI) = k^2$$ for a $$2 \times 2$$ matrix.

So $$\det\left(A^2 - \frac{1}{2}I\right) = \left(\sin^2\alpha - \frac{1}{2}\right)^2 = 0$$.

This gives $$\sin^2\alpha = \frac{1}{2}$$, so $$\sin\alpha = \pm \frac{1}{\sqrt{2}}$$.

Therefore $$\alpha = \frac{\pi}{4}$$ (among the given options), which matches Option C.