Step 1: Split the matrix into its block structure.
The matrix is block diagonal: a $1 \times 1$ block $[1]$ in the top left, and a $2 \times 2$ rotation-style block $\begin{bmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{bmatrix}$ in the bottom right. Eigenvalues of a block-diagonal matrix are just the eigenvalues of each block, combined.
Step 2: Find the eigenvalue of the $1\times1$ block.
The top-left block is simply the number $1$, so it contributes the eigenvalue $\gamma_1 = 1$ directly.
Step 3: Find the eigenvalues of the $2\times2$ block using its characteristic equation.
For $\begin{bmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{bmatrix}$, the characteristic polynomial is $$\lambda^{2} - (2\cos t)\lambda + (\cos^{2}t + \sin^{2}t) = 0$$ $$\lambda^{2} - (2\cos t)\lambda + 1 = 0$$ since $\cos^2 t + \sin^2 t = 1$. Solving this quadratic with the quadratic formula gives $\lambda = \cos t \pm i\sin t$, which by Euler's formula are $e^{it}$ and $e^{-it}$.
Step 4: Add up all three eigenvalues.
$$\gamma_1 + \gamma_2 + \gamma_3 = 1 + e^{it} + e^{-it} = 1 + 2\cos t$$ using the identity $e^{it} + e^{-it} = 2\cos t$. This confirms the same trace-based result reached from a different route.
Step 5: Solve for $t$.
$$1 + 2\cos t = 1 + \sqrt{2} \implies \cos t = \frac{1}{\sqrt{2}}$$ Within $[-\pi, \pi]$, the solutions are $t = \pi/4$ and $t = -\pi/4$, since these are exactly the angles whose cosine is $1/\sqrt{2}$.
Conclusion:
This matches option (C), $\{\pi/4, -\pi/4\}$; the pair involving $\pi/3$ would need $\cos t = 1/2$, which is a different value entirely.$$\boxed{t \in \{\pi/4, -\pi/4\}}$$