The maximum and minimum value of \[ a \cos \alpha + b \sin \alpha = \pm \sqrt{a^2 + b^2} \] Hence, the maximum and minimum value of \[ 3 \cos \alpha + 4 \sin \alpha = \pm 5 \] Given that \[ |k| = 5 \Rightarrow k = \pm 5 \] Taking \[ k = 5, \quad 3 \cos \alpha + 4 \sin \alpha = 5 \] or \[ \frac{3}{5} \cos \alpha + \frac{4}{5} \sin \alpha = 1 \] This is possible only when \[ \cos \alpha = \frac{3}{5} \quad \text{and} \quad \sin \alpha = \frac{4}{5} \] Similarly, when \[ 3 \cos \alpha + 4 \sin \alpha = -5 \] or \[ \frac{3}{5} \cos \alpha + \frac{4}{5} \sin \alpha = -1 \] This is possible only when \[ \cos \alpha = -\frac{3}{5} \quad \text{and} \quad \sin \alpha = -\frac{4}{5} \] Hence, there are two possible values of \(\alpha\), so there are two solutions.