6. Find the sum of the integers that belong to the set of values of the function $f(x)=\log _{2}(5 \cos 2 x+11)$ for $x \in[1,25(\operatorname{arctg}(1 / 3)) \cos (\pi+\arcsin (-0.6)) ; \operatorname{arctg} 2] \quad$ (10 points)
# Solution:

Since $\quad$ $\cos (\pi+\arcsin (-0.6))=\cos (\pi-\arcsin 0.6)=-\cos (\arcsin 0.6)=-0.8, \quad$ then $x \in[1.25(\operatorname{arctg}(1 / 3)) \cos (\pi+\arcsin (-0.6)) ; \operatorname{arctg} 2]=[-\operatorname{arctg}(1 / 3) ; \operatorname{arctg} 2]$ Therefore, $2 x \in[-2 \operatorname{arctg}(1 / 3) ; 2 \operatorname{arctg} 2]$ Since $\quad 0<\operatorname{arctg}(1 / 3)<\operatorname{arctg} 1=\pi / 4, \quad 0<2 \operatorname{arctg}(1 / 3)<\pi / 2$, $-\pi / 2<-2 \operatorname{arctg}(1 / 3)<0, \quad$ and $\quad \pi / 4<\operatorname{arctg} 2<\pi / 2, \quad \pi / 2<2 \operatorname{arctg} 2<\pi, \quad$ then $\cos 2 x \in[\cos (2 \operatorname{arctg} 2) ; 1] . \quad$ Using the formula $\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}, \quad$ we get $\cos (2 \operatorname{arctg} 2)=-0.6, \quad$ and $\quad \cos 2 x \in[-0.6 ; 1]$. From this, we have $5 \cos 2 x+11 \in[8 ; 16]$, and $f(x)=\log _{2}(5 \cos 2 x+11) \in\left[\log _{2} 8 ; \log _{2} 16\right]=[3 ; 4]$.

The interval $[3 ; 4]$ is the range of the function $f(x)=\log _{2}(5 \cos 2 x+11)$ for $x \in[1.25(\operatorname{arctg}(1 / 3)) \cos (\pi+\arcsin (-0.6)) ; \operatorname{arctg} 2]$.

The sum of the integers in the interval $[3 ; 4]$ is 7.

Answer: $E_{f}=[3 ; 4]$, the sum of the integers is 7.