Step-by-step Solution:

\[ \mathbf{a} \times (\mathbf{a} \times \mathbf{c}) + \mathbf{b} = 0 \] \[ \Rightarrow (\mathbf{a} \cdot \mathbf{c}) \mathbf{a} - (\mathbf{a} \cdot \mathbf{a}) \mathbf{c} = -\mathbf{b} \] \[ \Rightarrow \mathbf{c} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{a} = \mathbf{b} \quad (\because \mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 = 1) \] \[ \Rightarrow |\mathbf{c} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{a}|^2 = |\mathbf{b}|^2 \] \[ \Rightarrow |\mathbf{c}|^2 + |\mathbf{a} \cdot \mathbf{c}|^2 |\mathbf{a}|^2 - 2 (\mathbf{a} \cdot \mathbf{c}) (\mathbf{a} \cdot \mathbf{c}) = |\mathbf{b}|^2 \] If \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{c}\), we get: \[ 4 + (2 \cos \theta)^2 - 2 (2 \cos \theta)^2 = 1 \] \[ \Rightarrow 4 \cos^2 \theta = 3 \Rightarrow \cos \theta = \frac{\sqrt{3}}{2} \quad (\because \theta \text{ is acute}) \] \[ \Rightarrow \theta = \frac{\pi}{6} \]