Problem (Algebra):
Determine all pairs of angles $(x, y)$ with $0^{\circ} \leq x<180^{\circ}$ and $0^{\circ} \leq y<180^{\circ}$ that satisfy the following system of equations:

$$
\begin{aligned}
\log _{2}(\sin x \cos y) & =-\frac{3}{2} \\
\log _{2}\left(\frac{\sin x}{\cos y}\right) & =\frac{1}{2}
\end{aligned}
$$

Solution:
Using logarithm laws, the given equations are equivalent to

$$
\begin{aligned}
& \log _{2}(\sin x)+\log _{2}(\cos y)=-\frac{3}{2} \\
& \log _{2}(\sin x)-\log _{2}(\cos y)=\frac{1}{2}
\end{aligned}
$$

Adding these two equations, we obtain $2 \log _{2}(\sin x)=-1$ which gives $\log _{2}(\sin x)=-\frac{1}{2}$ and so $\sin x=2^{-1 / 2}=\frac{1}{2^{1 / 2}}=\frac{1}{\sqrt{2}}$.

Since $0^{\circ} \leq x<180^{\circ}$, then $x=45^{\circ}$ or $x=135^{\circ}$.

Since $\log _{2}(\sin x)+\log _{2}(\cos y)=-\frac{3}{2}$ and $\log _{2}(\sin x)=-\frac{1}{2}$, then $\log _{2}(\cos y)=-1$, which gives $\cos y=2^{-1}=\frac{1}{2}$.

Since $0^{\circ} \leq y<180^{\circ}$, then $y=60^{\circ}$.

Therefore, $(x, y)=\left(45^{\circ}, 60^{\circ}\right)$ or $(x, y)=\left(135^{\circ}, 60^{\circ}\right)$.

First, we note that $2^{1 / 2}=\sqrt{2}$ and $2^{-3 / 2}=\frac{1}{2^{3 / 2}}=\frac{1}{2^{1} 2^{1 / 2}}=\frac{1}{2 \sqrt{2}}$.

From the given equations, we obtain

$$
\begin{aligned}
\sin x \cos y & =2^{-3 / 2}=\frac{1}{2 \sqrt{2}} \\
\frac{\sin x}{\cos y} & =2^{1 / 2}=\sqrt{2}
\end{aligned}
$$

Multiplying these two equations together, we obtain $(\sin x)^{2}=\frac{1}{2}$ which gives $\sin x= \pm \frac{1}{\sqrt{2}}$.

Since $0^{\circ} \leq x<180^{\circ}$, it must be the case that $\sin x \geq 0$ and so $\sin x=\frac{1}{\sqrt{2}}$.

Since $0^{\circ} \leq x<180^{\circ}$, we obtain $x=45^{\circ}$ or $x=135^{\circ}$.

Since $\sin x \cos y=\frac{1}{2 \sqrt{2}}$ and $\sin x=\frac{1}{\sqrt{2}}$, we obtain $\cos y=\frac{1}{2}$.

Since $0^{\circ} \leq y<180^{\circ}$, then $y=60^{\circ}$.

Therefore, $(x, y)=\left(45^{\circ}, 60^{\circ}\right)$ or $(x, y)=\left(135^{\circ}, 60^{\circ}\right)$.

Final answer: $(45^{\circ}, 60^{\circ}),(135^{\circ}, 60^{\circ})$