\( \begin{array}{l}\text { Find one solution for the equation. Assume that all angles involved are acute angles. } \\ \sec \left(3 \beta+14^{\circ}\right)=\csc \left(2 \beta-4^{\circ}\right) \\ \beta=\square^{\circ} \\ \text { (Simplify your answer.) }\end{array} \)

To solve the equation \( \sec(3\beta + 14^{\circ}) = \csc(2\beta - 4^{\circ}) \), we can rewrite it in terms of sine and cosine. Recall that \( \sec(x) = \frac{1}{\cos(x)} \) and \( \csc(y) = \frac{1}{\sin(y)} \). Thus, we can rewrite the equation as: \[ \frac{1}{\cos(3\beta + 14^{\circ})} = \frac{1}{\sin(2\beta - 4^{\circ})} \] This leads to: \[ \sin(2\beta - 4^{\circ}) = \cos(3\beta + 14^{\circ}) \] Using the co-function identity \( \sin(x) = \cos(90^{\circ} - x) \), we can rewrite \(\cos(3\beta + 14^{\circ})\): \[ \sin(2\beta - 4^{\circ}) = \sin\left(90^{\circ} - (3\beta + 14^{\circ})\right) \] This simplifies to: \[ 2\beta - 4^{\circ} = 90^{\circ} - (3\beta + 14^{\circ}) + k \cdot 180^{\circ} \quad \text{or} \quad 2\beta - 4^{\circ} = 180^{\circ} - \left(90^{\circ} - (3\beta + 14^{\circ})\right) + k \cdot 180^{\circ} \] We'll simplify the first equation: \[ 2\beta - 4^{\circ} = 76^{\circ} - 3\beta + k \cdot 180^{\circ} \] Reorganizing gives us: \[ 5\beta = 76^{\circ} + 4^{\circ} + k \cdot 180^{\circ} \] \[ 5\beta = 80^{\circ} + k \cdot 180^{\circ} \] Dividing by 5, we have: \[ \beta = 16^{\circ} + k \cdot 36^{\circ} \] Assuming \( k = 0 \) for the acute angle, we find: \[ \beta = 16^{\circ} \] So one solution is: \(\beta = 16^{\circ}\)