6. Find the set of values of the function $f(x)=\sqrt{36-g^{2}(x)}$, where $g(x)=-8-2 \cos 8 x-4 \cos 4 x$.

#
# Solution:

$g(x)=-8-2 \cos 8 x-4 \cos 4 x=-8-4 \cos ^{2} 4 x+2-4 \cos 4 x=$

$-4 \cos ^{2} 4 x-4 \cos 4 x-6=-(2 \cos 4 x+1)^{2}-5$.

The function $t=2 \cos 4 x$ takes values $t \in[-2 ; 2]$. Consider the function $y=-5-(t+1)^{2}$, defined on the interval [-2; 2]. The graph of this function is a parabola with its vertex at the point $(-1 ;-5)$, with branches directed downwards. Thus, the maximum value of $y$ is -5, which is achieved at the point $t=-1$, the minimum value of the function is -14, which it takes at the point $t=2$, and the function takes all values from the interval $[-14 ;-5]$.

The function $f(x)=\sqrt{36-g^{2}(x)}$ is defined for those $x$ for which $g(x)$ takes values from the interval $[-6 ;-5]$. The set of values of the function $f(x)$ is the set $E_{f}=\lfloor 0 ; \sqrt{11}\rfloor$.

Answer: $E_{f}=\lfloor 0 ; \sqrt{11}\rfloor$.