4. In all other cases - $\mathbf{0}$ points.

## Task 2

## Maximum 15 points

Solve the equation $2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\sin \left(\frac{\pi}{4}(1+x)\right)$.

How many solutions of this equation satisfy the condition: $2000 \leq x \leq 3000$?
# Solution:

$\sin \left(\frac{\pi}{4}(1+x)\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$. The equation becomes

$2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$, i.e., we get problem 2 from option 1, the solution of which is:

$x=1+4 n, n \in Z . \quad 2000 \leq x \leq 3000,2000 \leq 1+4 n \leq 3000,499.75 \leq n \leq 749.75$.

Since $n$ is an integer, $500 \leq n \leq 749$. The number of integers $n$ is 250.

Answer: 250.

## Grading Criteria: