6. Find the sum of the integers that belong to the set of values of the function $f(x)=\log _{3}(10 \cos 2 x+17)$ for $x \in[1,25(\operatorname{arctg} 0.25) \cos (\pi-\arcsin (-0.6)) ; \operatorname{arctg} 3] . \quad(10$ points $)$
# Solution:

Since $\quad$ $\cos (\pi-\arcsin (-0.6))=\cos (\pi+\arcsin 0.6)=-\cos (\arcsin 0.6)=-0.8, \quad$ then $x \in[1.25(\operatorname{arctg} 0.25) \cos (\pi-\arcsin (-0.6)) ; \operatorname{arctg} 3]=[-\operatorname{arctg} 0.25 ; \operatorname{arctg} 3] \quad$ Therefore, $2 x \in[-2 \operatorname{arctg} 0.25 ; 2 \operatorname{arctg} 3]$ Since $\quad 0<\operatorname{arctg} 0.25<\operatorname{arctg} 1=\pi / 4, \quad 0<2 \operatorname{arctg} 0.25<\pi / 2$, $-\pi / 2<-2 \operatorname{arctg} 0.25<0, \quad$ and $\quad \pi / 4<\operatorname{arctg} 3<\pi / 2, \quad \pi / 2<2 \operatorname{arctg} 3<\pi, \quad$ then $\cos 2 x \in[\cos (2 \operatorname{arctg} 3) ; 1] . \quad$ Using $\quad$ the formula $\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}, \quad$ we get $\cos (2 \operatorname{arctg} 3)=-0.8, \quad$ and $\cos 2 x \in[-0.8 ; 1]$. From this, we have $10 \cos 2 x+17 \in[9 ; 27]$, and $f(x)=\log _{3}(10 \cos 2 x+17) \in\left[\log _{3} 9 ; \log _{3} 27\right]=[2 ; 3]$.

The interval $[2 ; 3]$ is the range of the function $f(x)=\log _{3}(10 \cos 2 x+17)$ for $x \in[1.25(\operatorname{arctg} 0.25) \cos (\pi-\arcsin (-0.6)) ; \operatorname{arctg} 3]$.

The sum of the integers in the interval $[2 ; 3]$ is 5.

Answer: $E_{f}=[2 ; 3]$, the sum of the integers is 5.