6. Find the sum of the integers that belong to the set of values of the function $f(x)=\log _{3}(40 \cos 2 x+41)$ for $x \in[(5 / 3)(\operatorname{arctg}(1 / 5)) \cos (\pi-\arcsin (-0.8)) ; \operatorname{arctg} 3]$ (10 points)
Solution: Since $\cos (\pi-\arcsin (-0.8))=\cos (\pi+\arcsin 0.8)=-\cos (\arcsin 0.8)=-0.6$, then $x \in[(5 / 3)(\operatorname{arctg}(1 / 5)) \cos (\pi-\arcsin (-0.8)) ; \operatorname{arctg} 3]=[-\operatorname{arctg}(1 / 5) ; \operatorname{arctg} 3]$. $2 x \in[-2 \operatorname{arctg}(1 / 5) ; 2 \operatorname{arctg} 3]$. Since $\quad 0<\operatorname{arctg}(1 / 5)<\operatorname{arctg} 1=\pi / 4, \quad 0<2 \operatorname{arctg}(1 / 5)<\pi / 2$, $-\pi / 2<-2 \operatorname{arctg}(1 / 5)<0$, and also $\pi / 4<\operatorname{arctg} 3<\pi / 2, \pi / 2<2 \operatorname{arctg} 3<\pi$, then $\cos 2 x \in[\cos (2 \operatorname{arctg} 3) ; 1]$. Using the formula $\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}$, we get $\cos (2 \operatorname{arctg} 3)=-0.8$, and $\cos 2 x \in[-0.8 ; 1]$. From this, we have $40 \cos 2 x+41 \in[9 ; 81]$, and $f(x)=\log _{3}(40 \cos 2 x+41) \in\left[\log _{3} 9 ; \log _{3} 81\right]=[2 ; 4]$.

The interval $[2 ; 4]$ is the range of the function $f(x)=\log _{3}(40 \cos 2 x+41)$ for $x \in[(5 / 3)(\operatorname{arctg}(1 / 5)) \cos (\pi-\arcsin (-0.8)) ; \operatorname{arctg} 3]$. The sum of the integers in the interval $[2 ; 4]$ is 9.

Answer: $E_{f}=[2 ; 4]$, the sum of the integers is 9.