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 $$ y=\square \cdot \square^{\circ}, k \in \mathbb{Z} $$

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We start with
$$
\frac{1-\sin ^{2}4y}{5\sin\left(90^{\circ}+4y\right)}=-0.06.
$$
Recall the Pythagorean identity
$$
1-\sin^2\theta=\cos^2\theta.
$$
Thus, the numerator becomes
$$
\cos^2(4y).
$$

Also, use the cofunction identity in degrees:
$$
\sin(90^{\circ}+4y)=\cos(4y).
$$
Substitute these into the equation:
$$
\frac{\cos^2(4y)}{5\cos(4y)}=-0.06.
$$

Assuming $$\cos(4y)\neq0$$, we can cancel one $$\cos(4y)$$:
$$
\frac{\cos(4y)}{5}=-0.06.
$$
Multiply both sides by 5:
$$
\cos(4y)=-0.30.
$$

The general solution for an equation of the form
$$
\cos \theta = -0.30
$$
is given by
$$
\theta = \pm \arccos(-0.30)+360^{\circ}k,\quad k\in\mathbb{Z}.
$$

Here, $$\theta=4y$$ and
$$
\arccos(-0.30) \approx 107.74^{\circ}\quad (\text{rounded to two decimal places}).
$$
Thus, we have
$$
4y=\pm 107.74^{\circ}+360^{\circ}k,\quad k\in\mathbb{Z}.
$$

Divide by 4 to solve for $$y$$:
$$
y=\pm\frac{107.74^{\circ}}{4}+\frac{360^{\circ}}{4}k,\quad k\in\mathbb{Z}.
$$
Simplify the fractions:
$$
\frac{107.74^{\circ}}{4}\approx 26.94^{\circ}\quad \text{and}\quad \frac{360^{\circ}}{4}=90^{\circ}.
$$
Hence, the solutions are
$$
y=26.94^{\circ}+90^{\circ}k,\quad k\in\mathbb{Z},
$$
or
$$
y=-26.94^{\circ}+90^{\circ}k,\quad k\in\mathbb{Z}.
$$

Equivalently, this can be written in the combined form:
$$
y=90^{\circ}k\pm26.94^{\circ},\quad k\in\mathbb{Z}.
$$
The general solution is: $$ y = 90^\circ \cdot k \pm 26.94^\circ, \quad k \in \mathbb{Z} $$

Hence, determine the general solution of

INSTRUCTION: Round your answer to two decimal places, if necessary.
Answer:

or

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Hence, determine the general solution of INSTRUCTION: Round your answer to two decimal places, if necessary. Answer: or