1. Solve the equation: $\sin \left(\frac{\pi}{2} \cos x\right)=\cos \left(\frac{\pi}{2} \sin x\right)$.
Answer: $\frac{\pi}{2}+\pi k, \quad k \in Z ; \quad 2 \pi k, \quad k \in Z$.

## Solution:

Using the reduction formula, we transform the right side of the equation:

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

Equality of sines is possible in one of the following cases:

Case 1: $\frac{\pi}{2} \cos x=\frac{\pi}{2}-\frac{\pi}{2} \sin x+2 \pi n$ (for integer $n$).

We transform this to: $\cos x+\sin x=1+4 n$.

For $n \geq 1$, we get $\cos x+\sin x \geq 5$. For $n \leq -1$, we get $\cos x+\sin x \leq -3$.

Both of these cases are impossible (even a rough estimate of $\cos x+\sin x$ does not exceed 2 in absolute value). Therefore, $n=0$, and the equation reduces to: $\cos x+\sin x=1$.

We square everything and transform it to: $2 \cos x \sin x=0$.

Then there are two possibilities:

(1) $\cos x=0$, then from the equality $\cos x+\sin x=1$ we get $\sin x=1$ - this satisfies the original equation: $\sin \left(\frac{\pi}{2} \cdot 0\right)=\cos \left(\frac{\pi}{2} \cdot 1\right)$, so all such $x$ (for which $\cos x=0$ and $\sin x=1$) are valid - these are the values $x=\frac{\pi}{2}+2 \pi k, \quad k \in Z$.

(2) $\sin x=0$, then from the equality $\cos x+\sin x=1$ we get $\cos x=1$ - this satisfies the original equation: $\sin \left(\frac{\pi}{2} \cdot 1\right)=\cos \left(\frac{\pi}{2} \cdot 0\right)$, all such $x$ (for which $\sin x=0$ and $\cos x=1$) are valid - these are the values $x=2 \pi k, \quad k \in Z$.

Case 2: $\frac{\pi}{2} \cos x=\pi-\left(\frac{\pi}{2}-\frac{\pi}{2} \sin x\right)+2 \pi n$ (for integer $n$).

We transform this to: $\cos x-\sin x=1-4 n$.

Similarly to the previous case, we conclude that the equality is possible only for $n=0$, and the equation reduces to $\cos x-\sin x=1$.

We solve it in the same way as above, obtaining two series of solutions: $x=-\frac{\pi}{2}+2 \pi k, k \in Z$ and $x=2 \pi k, \quad k \in Z$ (duplicates the previously obtained solution).

## Criteria:

A fully correct answer without a correct justification - 2 points.

Non-equivalent transitions-consequences without verification (why we don't get extraneous roots) should not be more than 3 points (if the answer is correct).

No explanations why the equation reduces to one of the cases $\cos x+\sin x=1$ or $\cos x-\sin x=1$ - should not be more than 3 points (if the answer is correct).