Hence, determine the general solution of \[ \frac{1-\sin ^{2} 4 y}{5 \sin \left(90^{\circ}+4 y\right)}=-0,06 \] INSTRUCTION: Round your answer to two decimal places, if necessary. Answer: \( y=\square \cdot k \cdot, k \in \mathbb{Z} \) or \( y=\square \cdot \square^{\circ}, k \in \mathbb{Z} \)

To solve the equation \[ \frac{1-\sin^{2} 4y}{5 \sin(90^{\circ}+4y)}=-0.06, \] we first simplify \(1-\sin^{2} 4y\) to \(\cos^{2} 4y\). The sine function has the identity \(\sin(90^{\circ} + x) = \cos x\), thus \(\sin(90^{\circ} + 4y) = \cos(4y)\). The equation then becomes \[ \frac{\cos^{2} 4y}{5 \cos(4y)} = -0.06. \] This simplifies to \[ \frac{\cos(4y)}{5} = -0.06 \quad \Rightarrow \quad \cos(4y) = -0.30. \] The general solution for \(\cos(θ) = x\) is given by \[ θ = \cos^{-1}(x) + 360^{\circ} k \text{ or } θ = -\cos^{-1}(x) + 360^{\circ} k. \] Thus, we solve for \(4y\): 1. \(4y = \cos^{-1}(-0.30) + 360^{\circ} k, \) 2. \(4y = 360^{\circ} - \cos^{-1}(-0.30) + 360^{\circ} k. \] Calculating \( \cos^{-1}(-0.30) \) gives approximately \(108.21^{\circ}\). Now, substituting back to solve for \(y\): 1. \(y = \frac{108.21^{\circ}}{4} + 90^{\circ} k, \) 2. \(y = \frac{360^{\circ} - 108.21^{\circ}}{4} + 90^{\circ} k. This results in two values: 1. \(y \approx 27.05^{\circ} + 90^{\circ} k, \) 2. \(y \approx 62.20^{\circ} + 90^{\circ} k. So, the general solution is: \[ y \approx 27.05^{\circ} + 90^{\circ} k \text{ or } y \approx 62.20^{\circ} + 90^{\circ} k, \quad k \in \mathbb{Z}. \] Final rounded answers are: \(y=27.05^{\circ} + 90^{\circ} k\) or \(y=62.20^{\circ} + 90^{\circ} k, k \in \mathbb{Z}\).