3A. In the set of real numbers, solve the equation

$$
\frac{a \sin x + b}{b \cos x + a} = \frac{a \cos x + b}{b \sin x + a}, a, b \in \mathbb{R} \text{ and } a, b > 0
$$
Solution. From the equation, it follows that $b \cos x+a \neq 0$ and $b \sin x+a \neq 0$. Multiplying the equation by $(b \cos x+a)(b \sin x+a)$, we obtain the equation

$$
(a \sin x+b)(b \sin x+a)=(a \cos x+b)(b \cos x+a)
$$

which is equivalent to the original equation.

By multiplying and appropriately grouping, we obtain the equation

$$
(\sin x-\cos x)\left(a^{2}+b^{2}+a b(\sin x+\cos x)\right)=0
$$

Thus,

$$
\sin x-\cos x=0 \text { or } a^{2}+b^{2}+a b(\sin x+\cos x)=0 \text {. }
$$

From $\sin x-\cos x=0$ we obtain that $x=\frac{\pi}{4}+k \pi, k \in \mathbb{Z}$ are solutions to the original equation.

From $a^{2}+b^{2}+a b(\sin x+\cos x)=0$ and $a b>0$ we obtain that

$$
\sin x+\cos x=-\frac{a^{2}+b^{2}}{a b}
$$

Using that $a^{2}+b^{2} \geq 2 a b$ we obtain $\sin x+\cos x=-\frac{a^{2}+b^{2}}{a b} \leq-\frac{2 a b}{a b}=-2$. Multiplying this inequality by $\frac{\sqrt{2}}{2}$ we obtain

$$
\frac{\sqrt{2}}{2} \sin x+\frac{\sqrt{2}}{2} \cos x=-2 \frac{\sqrt{2}}{2}, \text { i.e., } \sin \left(\frac{\pi}{4}+x\right)=-\sqrt{2}
$$

which is impossible. Therefore, the equation $a^{2}+b^{2}+a b(\sin x+\cos x)=0$ has no solution.

Accordingly, all solutions to the original equation are $x=\frac{\pi}{4}+k \pi, k \in \mathbb{Z}$

Note. The conclusion that the equation $a^{2}+b^{2}+a b(\sin x+\cos x)=0$ has no solution can also be derived as follows:

It is clear that $\sin x+\cos x \geq-2$.

If $\sin x+\cos x=-2$, due to $\sin x \geq-1$ and $\cos x \geq-1$, we obtain that $\sin x=-1$ and $\cos x=-1$, so $\sin ^{2} x+\cos ^{2} x=2$ which is impossible.

Therefore, $\sin x+\cos x>-2$.

Now we have $a^{2}+b^{2}+a b(\sin x+\cos x)>a^{2}+b^{2}-2 a b=(a-b)^{2} \geq 0$, so the equation $a^{2}+b^{2}+a b(\sin x+\cos x)=0$ has no solution.