Solve the equation $\sin (\pi \cos x)=\cos (\pi \sin x)$.
1. Solution. Based on the identity $\cos \alpha=\sin \left(\frac{\pi}{2}+\alpha\right)$, the equation to be solved can be written as:

$$
\sin (\pi \cdot \cos x)=\sin \left(\frac{\pi}{2}+\pi \sin x\right)
$$

This holds if and only if

$$
\pi \cdot \cos x=\frac{\pi}{2}+\pi \sin x+2 \cdot k \cdot \pi(1) \text { or } \pi \cdot \cos x=\pi-\left(\frac{\pi}{2}+\pi \sin x\right)+2 \cdot k \cdot \pi \quad(k \in \mathbf{Z}) \cdot(2)
$$

We need to solve the two equations connected by "or" separately, so the parameter taken in both can be denoted by $k$. Equations (1) and (2) can be written as:

$$
\cos x-2 k-\frac{1}{2}=\sin x,(3) \cos x-2 k-\frac{1}{2}=-\sin x,(4)
$$

and the disjunction of these two equations is equivalent to the original equation. Squaring equation (3) and using the identity $\sin ^{2} x=$ $1-\cos ^{2} x$:

$$
\cos ^{2} x+\left(2 k+\frac{1}{2}\right)^{2}-2\left(2 k+\frac{1}{2}\right) \cos x=1-\cos ^{2} x
$$

From this,

$$
2 \cdot \cos ^{2} x-(4 k+1) \cos x+\left(4 k^{2}+2 k-\frac{3}{4}\right)=0
$$

It is obvious that squaring equation (4) also leads to (5), so the disjunction of (3) and (4) is equivalent to (5).

The discriminant of (5) is: $D=(4 k+1)^{2}-4 \cdot 2\left(4 k^{2}+2 k-\frac{3}{4}\right)=-16 k^{2}-8 k+7$, which is non-negative only when $k=0$, in which case $D=7$. Therefore, the real solutions of (5) and thus of the original equation are obtained from the equations $\cos x=\frac{4 k+1 \pm \sqrt{7}}{4}=\frac{1 \pm \sqrt{7}}{4}$. This gives $\cos x=0.9114$ or $\cos x=-0.4114$, from which $x= \pm 0.4240+2 m \cdot \pi ;$ or $x= \pm 1.995+2 m \cdot \pi$ $(m \in \mathbf{Z})$ are all the solutions. Due to the mentioned equivalences, these solutions satisfy the original equation.

$$
\text { Makai Márton (Debrecen, Fazekas M. Gimn., III. o. t.) }
$$

Second solution. From equation (3) of the first solution, $\cos x-\sin x=2 k+\frac{1}{2}$. Multiplying this by $\frac{1}{\sqrt{2}}$:

$$
\frac{1}{\sqrt{2}} \cdot \cos x-\frac{1}{\sqrt{2}} \cdot \sin x=\frac{4 k+1}{2 \cdot \sqrt{2}}, \sin \frac{\pi}{4} \cdot \cos x-\cos \frac{\pi}{4} \cdot \sin x=\frac{4 k+1}{2 \cdot \sqrt{2}}, \text{ so } \sin \left(\frac{\pi}{4}-x\right)=\frac{4 k+1}{2 \cdot \sqrt{2}} \cdot(6)
$$

This equation has a solution only if $\left|\frac{4 k+1}{2 \sqrt{2}}\right| \leq 1$, which implies that $k=0$. Then from (6):

$$
\sin \left(\frac{\pi}{4}-x\right)=\frac{1}{2 \sqrt{2}}
$$

Similarly, from (4):

$$
\sin \left(\frac{\pi}{4}+x\right)=\frac{1}{2 \cdot \sqrt{2}}
$$

From these two equations, we obtain the already known solutions.

Lovász Zoltán, (Bonyhád, Perczel Mór Közg. Szki., IV. o.t.)

Note: Lovász Zoltán solved the equation $\sin (a \cdot \cos x)=\cos (b \cdot \sin x)$ using the method of the second solution.