## Task 35/77

How many non-negative real roots does the function

$$
y=f(x)=x-1978 \sin \pi x
$$

have?
If we decompose the function $y=f(x)=x-1978 \sin \pi x$ into the two partial functions $y_{1}=f_{1}(x)=x$ and $y_{2}=f_{2}(x)=1978 \sin \pi x$, the problem boils down to how often $y_{1}=y_{2}$.

If we imagine the two functions $f_{1}$ and $f_{2}$ graphically, $f_{1}$ provides a straight line with a $45^{\circ}$ inclination passing through the origin, while $f_{2}$ represents a sine curve with an amplitude of 1978.

It is obvious that the straight line intersects the sine curve exactly twice in each period, provided that $0 \leq x \leq 1978$.

For $x>1978$, there are no intersection points, since $y_{1}>1978, y_{2} \leq 1978$.

Thus, it suffices to determine the number $n$ of periods of $f_{2}$ in the interval $[0 ; 1978]$ and double this number. Since the sine function has a period length of $2 \pi$, the period length of $f_{2}(x)$ is 2. The number $k=2 n$ of non-negative roots of $y=f(x)=x-1978 \sin \pi x$ is therefore

$$
k=2 n=\frac{1978-0}{2} \cdot 2=1978
$$

(the first of which lies at $x_{0}=0$).