5.1.1. A particle moves in the plane $O x y$ of a rectangular coordinate system according to the law (here $x, y-$ are the coordinates of the point in meters, $t$ - time in seconds):

$$
x(t)=3+\sin t \cdot \cos t-\sin t-\cos t ; y(t)=1
$$

In the same plane, a light ray constantly emanates from the origin according to the law $y=c x$, where $c-$ is a positive dimensionless constant. For what values of $c$ does there exist at least one moment in time when the particle is illuminated by the ray?
Solution.

Let's make the substitution of variables $z=\sin t+\cos t$. Then $z^{2}=1+2 \sin t \cos t$, and the dependence of $x$ on $t$ takes the form

$$
x(t)=3+\frac{z^{2}-1}{2}-z, \quad z=\sin t+\cos t, \quad |z| \leqslant \sqrt{2}
$$

The function $f(z)=3+\frac{z^{2}-1}{2}-z=\frac{z^{2}}{2}-z+\frac{5}{2}=\frac{1}{2}(z-1)^{2}+2$ is a parabola with vertex $z_{0}=1$, having a minimum at this vertex, which is 2.

Since $|z| \leqslant \sqrt{2}$, the minimum value of the function will be at the vertex, and the maximum value will be at $z=-\sqrt{2}$. We obtain that the range of this function is the interval $\left[2 ; \frac{7}{2}+\sqrt{2}\right]$.

Thus, the particle at different moments of time passes through all points of the segment $A B$, where $A(2 ; 1)$, $B\left(\frac{7}{2}+\sqrt{2} ; 1\right)$.

The ray given in the condition passes through the point $A$ when $c=\frac{1}{2}$, and through the point $B$ when

$c=\frac{1}{\frac{7}{2}+\sqrt{2}}=\frac{2}{7+2 \sqrt{2}}=\frac{2(7-2 \sqrt{2})}{41}$.

For all intermediate values of $c$, there will be moments of time when the ray illuminates the particle. Thus, .

Answer: $c \in\left[\frac{2(7-2 \sqrt{2})}{41} ; \frac{1}{2}\right]$