Step 1: Compute the determinant of $A$ to check invertibility.
Given $A = \begin{bmatrix}\sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha\end{bmatrix}$. Then $|A| = \sin^2\alpha + \cos^2\alpha = 1$. Since $|A|=1 \neq 0$ for all $\alpha$, the matrix $A$ is always invertible.
Step 2: Find the inverse $A^{-1}$.
For a $2\times2$ matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ with determinant $1$, the inverse is $\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$. So: \[A^{-1} = \begin{bmatrix}\sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha\end{bmatrix}\]
Step 3: Compute $A + A^{-1}$.
\[A+A^{-1} = \begin{bmatrix}2\sin\alpha & 0 \\ 0 & 2\sin\alpha\end{bmatrix}\] The off-diagonal entries cancel ($-\cos\alpha + \cos\alpha = 0$), and the diagonal entries add.
Step 4: Apply the condition $A + A^{-1} = I$.
Setting $\begin{bmatrix}2\sin\alpha & 0 \\ 0 & 2\sin\alpha\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$, we compare diagonal entries and get $2\sin\alpha = 1$, so $\sin\alpha = 1/2$.
Step 5: Solve $\sin\alpha = 1/2$ for the principal value.
We know $\sin(\pi/6) = 1/2$. Among the answer choices, $\alpha = \pi/6$ is option 3 (the correct one). The other values: $\sin(0)=0$, $\sin(\pi/3)=\sqrt{3}/2$, $\sin(\pi/4)=1/\sqrt{2}$ -- none equal $1/2$.
Step 6: Verify and state the final answer.
With $\alpha = \pi/6$: $A + A^{-1} = \begin{bmatrix}1&0\\0&1\end{bmatrix} = I$. Confirmed. \[ \boxed{\alpha = \frac{\pi}{6}} \]