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

To solve the equation: \[ \sec(3\beta + 14^\circ) = \csc(2\beta - 4^\circ) \] given that all angles involved are acute, we can proceed as follows: 1. **Rewrite the Equation Using Trigonometric Identities:** \[ \sec(\theta) = \frac{1}{\cos(\theta)} \quad \text{and} \quad \csc(\phi) = \frac{1}{\sin(\phi)} \] Therefore, \[ \frac{1}{\cos(3\beta + 14^\circ)} = \frac{1}{\sin(2\beta - 4^\circ)} \] This simplifies to: \[ \sin(2\beta - 4^\circ) = \cos(3\beta + 14^\circ) \] 2. **Convert \(\cos\) to \(\sin\):** Recall that \(\cos(\theta) = \sin(90^\circ - \theta)\). Applying this: \[ \sin(2\beta - 4^\circ) = \sin(90^\circ - (3\beta + 14^\circ)) = \sin(76^\circ - 3\beta) \] 3. **Set Up the Equations:** The equality \(\sin(A) = \sin(B)\) holds if: \[ A = B + 360^\circ k \quad \text{or} \quad A = 180^\circ - B + 360^\circ k \quad \text{for some integer } k \] Applying this: - **First Case:** \[ 2\beta - 4^\circ = 76^\circ - 3\beta \] Solving for \(\beta\): \[ 5\beta = 80^\circ \implies \beta = 16^\circ \] - **Second Case:** \[ 2\beta - 4^\circ = 180^\circ - (76^\circ - 3\beta) \implies 2\beta - 4^\circ = 104^\circ + 3\beta \] Solving for \(\beta\): \[ -\beta = 108^\circ \implies \beta = -108^\circ \] This solution is negative and does not satisfy the acute angle condition. 4. **Verify the Solution:** For \(\beta = 16^\circ\): - \(3\beta + 14^\circ = 48^\circ + 14^\circ = 62^\circ\) (acute) - \(2\beta - 4^\circ = 32^\circ - 4^\circ = 28^\circ\) (acute) Both angles are indeed acute, satisfying the given conditions. **Final Answer:** \(\boxed{16^\circ}\)