5. Find all values of the parameter $b$, for which for any value of the parameter $a \in[-2 ; 1]$ the inequality $a^{2}+b^{2}-\sin ^{2} 2 x-2(a+b) \cos 2 x-2>0$ is not satisfied for at least one value of $x$.

(20 points)
Solution: Let's make the substitution $y=\cos 2 x, \quad y \in[-1 ; 1]$. We get $a^{2}+b^{2}-1+y^{2}-2(a+b) y-2>0$, or $y^{2}-2(a+b) y+a^{2}+b^{2}-3>0$. We need to determine for which values of $a$ and $b$ the inequality $y^{2}-2(a+b) y+a^{2}+b^{2}-3>0$ holds for any $y \in[-1 ; 1]$. Consider the function $f(y)=y^{2}-2(a+b) y+a^{2}+b^{2}-3$. Its graph is a parabola with branches directed upwards, with the vertex at the point with abscissa $y_{0}=a+b$.

1) $\left\{\begin{array}{l}a+b \leq-1 \\ f(-1)>0\end{array}\right.$ $a+b \leq-1$,
$1)^{2}+(b+1)^{2}>4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f8bb21871417ffa2deg-3.jpg?height=274&width=237&top_left_y=1205&top_left_x=1321)
2) $\left\{\begin{array}{c}-10,\end{array} \quad\left\{\begin{array}{c}-10$ is not satisfied for at least one $y \in[-1 ; 1]$.

The points of intersection of the hyperbola $a b=-1.5$ and the line $a+b=1$ are found by solving the equation $t^{2}-t-1.5=0$. We get the points $((1 \pm \sqrt{7}) / 2 ;(1 \mp \sqrt{7}) / 2)$. The circle $(a-1)^{2}+(b-1)^{2}=4$ intersects the line $a+b=1$ at the same points (this can be verified by substitution). Similar reasoning is applied to the second

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f8bb21871417ffa2deg-3.jpg?height=788&width=834&top_left_y=2073&top_left_x=1199)
circle and the line. In the end, the points for which the inequality $y^{2}-2(a+b) y+a^{2}+b^{2}-3>0$ is not satisfied for at least one $y \in[-1 ; 1]$ form a closed region, the boundary of which consists of the graphs of two circles and a hyperbola, with the boundary included. To solve the problem, we need to find such values of $b$ for which the points $(a ; b)$ fall into the resulting region for any $a \in[-2 ; 1]$. These values of $b$ form the interval $\left[b_{1} ; b_{2}\right]$. The lower bound $b_{1}$ is found by substituting $a=1$ into the equation of the hyperbola. We get $b_{1}=-1.5$. The upper bound $b_{2}$ is found by substituting $a=-2$ into the equation of the circle $(a+1)^{2}+(b+1)^{2}=4$. We get $b_{2}=\sqrt{3}-1$.

Answer: $[-1.5 ; \sqrt{3}-1]$.