## Task A-3.1.

Determine all real numbers $a$ such that the inequality

$$
\cos (2 x)+2 a \cos x+7 \geqslant 0
$$

holds for all real numbers $x$.
## First Solution.

The expression on the left side can be written as

$$
\cos (2 x)+2 a \cos x+7=2 \cos ^{2} x-1+2 a \cos x+7=2\left(\cos ^{2} x+a \cos x+3\right)
$$

In the above expression, we can introduce the substitution $t=\cos x$. Since the function cos takes values between -1 and 1, we conclude that the problem becomes to determine all $a \in \mathbb{R}$ such that

$$
t^{2}+a t+3 \geqslant 0, \quad \text { for all } \quad t \in[-1,1]
$$

The abscissa of the vertex of the quadratic function is $t_{0}=-\frac{a}{2}$.

We consider three cases, depending on the position of the vertex relative to -1 and 1.

In the first case, $t_{0} \in[-1,1]$, i.e., when $a \in[-2,2]$, the quadratic function on the interval $[-1,1]$ attains its minimum at the vertex, so the non-negativity of the quadratic function on $[-1,1]$ is equivalent to the non-negativity of the quadratic function at $t_{0}$. It must hold that

$$
t_{0}^{2}+a t_{0}+3 \geqslant 0 \Longleftrightarrow \frac{-a^{2}}{4}+3 \geqslant 0 \Longleftrightarrow a^{2} \leqslant 12
$$

which holds for all $a \in[-2,2]$.

In the second case, $t_{0} < -1$ (i.e., when $a > 2$), the quadratic function on the interval $[-1,1]$ is increasing, so it attains its minimum at the left endpoint of the interval. Therefore, it must hold that

$$
0 \leqslant(-1)^{2}+a(-1)+3=4-a
$$

or $a \in\langle 2,4]$.

In the case when $t_{0}>1$ (i.e., when $a<-2$), we similarly conclude that the function on the interval $[-1,1]$ attains its minimum at the right endpoint of the interval, so it must hold that

$$
0 \leqslant 1^{2}+a \cdot 1+3=4+a
$$

from which $a \in[-4,-2\rangle$.

Finally, the set of sought real numbers $a$ is the interval

$$
[-4,-2\rangle \cup[-2,2] \cup\langle 2,4]=[-4,4]
$$