## Task B-3.4.

How many solutions does the equation $\frac{1}{2} \sin 2 x + 1 = \sin x + \cos x$ have on the interval $[0, 2020 \pi]$?
## Solution.

If we apply the formula for the sine of a double number, the equation becomes

$$
\sin x \cos x+1=\sin x+\cos x
$$

If we move all terms to the left side and factorize, we get

$$
\begin{gathered}
\sin x \cos x-\sin x-\cos x+1=0 \\
\sin x(\cos x-1)-(\cos x-1)=0 \\
\quad(\sin x-1)(\cos x-1)=0
\end{gathered}
$$

This means that $\sin x=1$ or $\cos x=1$.

The solutions to the first equation are $x_{1}=\frac{\pi}{2}+2 k \pi, k \in \mathbb{Z}$,

and the second $x_{2}=2 m \pi, m \in \mathbb{Z}$.

On the interval $[0,2 \pi\rangle$ the given equation has 2 solutions $x=0$ and $x=\frac{\pi}{2}$,

so due to periodicity, on the interval $[0,2020 \pi\rangle$ it will have $2 \cdot 1010=2020$ solutions, and on the interval $[0,2020 \pi]$ exactly 2021 solutions.