6. Find the set of values of the function

$$
f(x)=\sqrt{g^{2}(x)-245}, \text { where } g(x)=15-2 \cos 2 x-4 \sin x
$$
# Solution:

$$
f(x)=\sqrt{g^{2}(x)-245}, \text { where } g(x)=15-2 \cos 2 x-4 \sin x
$$

Let's find the range of the function $z=g(x)=15-2 \cos 2 x-4 \sin x$. The function $g(x)$ is defined for all real numbers. We will make a substitution. Let $t=\sin x$. Then $z=13+4 t^{2}-4 t=12+(2 t-1)^{2}$ for $t \in[-1 ; 1]$, and $E_{g}=[12 ; 21]$. The function $y=f(x)=\sqrt{g^{2}(x)-245}$ has the same range as the function $y=\sqrt{z^{2}-245}$ for $z \in[\sqrt{245} ; 21]$, since the function $y=\sqrt{z^{2}-245}$ is defined for $z \in(-\infty ;-\sqrt{245}] \cup[\sqrt{245} ;+\infty)$, and $z=g(x)$ takes all values from the interval $[12 ; 21]$.

The minimum value of the function $y=\sqrt{z^{2}-245}$ on the interval $[\sqrt{245} ; 21]$ is 0, and the maximum value is 14. Therefore, $E_{f}=[0 ; 14]$. Answer: $E_{f}=[0 ; 14]$.