4. In all other cases - o points.

## Task 2

Maximum 15 points

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

How many solutions of this equation satisfy the condition:

$0 \leq x \leq 2020 ?$
Solution. Let $t=\frac{\pi x}{4}$. Then the equation takes the form

$2 \sqrt{2} \sin ^{3} t=\cos \left(\frac{\pi}{4}-t\right)$.

$2 \sqrt{2} \sin ^{3} t=\cos \frac{\pi}{4} \cos t+\sin \frac{\pi}{4} \sin t$

$2 \sqrt{2} \sin ^{3} t=\frac{\sqrt{2}}{2} \cos t+\frac{\sqrt{2}}{2} \sin t$

$4 \sin ^{3} t=\cos t+\sin t ; 4 \sin ^{3} t-\sin t=\cos t$

Case 1: $\sin t=0 \rightarrow \cos t=0 \rightarrow \emptyset$

Case 2: $\sin t \neq 0$. Dividing both sides of the equation by $\sin t$, we get:

$4 \sin ^{2} t-1=\cot t$. Considering that $\sin ^{2} t=\frac{1}{1+\cot ^{2} t}$ and denoting $y=\cot t$, we obtain

$\frac{4}{1+y^{2}}-1=y$, or after transformation $y^{3}+y^{2}+y-3=0$

$(y-1)\left(y^{2}+2 y+3\right)=0$, the only root of which is $y=1$.

Making reverse substitutions: $\cot t=1 \rightarrow t=\frac{\pi}{4}+\pi n, n \in \mathbb{Z}$.

$\frac{\pi x}{4}=\frac{\pi}{4}+\pi n, n \in \mathbb{Z}$, from which $x=1+4 n, n \in \mathbb{Z}$.

$0 \leq x \leq 2020 ; 0 \leq 1+4 n \leq 2020, -0.25 \leq n \leq 504.75$.

Since $n$ is an integer, $0 \leq n \leq 504$. The number of integers $n$ is 505.

Answer: 505.

## Evaluation Criteria: