6. Find the set of values of the function $f(x)=\sqrt{1-g^{2}(x)}$, where $g(x)=\frac{\cos 6 x+2 \sin ^{2} 3 x}{2-2 \cos 3 x}$.
Solution: $\quad g(x)=\frac{\cos 6 x+2 \sin ^{2} 3 x}{2-2 \cos 3 x}=\frac{1}{2-2 \cos 3 x}$.

The function $t=\cos 3 x$ takes values $t \in[-1 ; 1]$. Consider the function $y=\frac{1}{2-2 t}$, defined on the half-interval $[-1 ; 1)$. The graph of this function is a hyperbola with
asymptotes $t=1$ and $y=0$. The function $y=\frac{1}{2-2 t}$ is unbounded and increasing on the interval $[-1 ; 1)$. Thus, the minimum value of $y$ is $\frac{1}{4}$, which is achieved at the point $t=-1$, and the function takes all values from the interval $\left[\frac{1}{4} ; \infty\right)$.

The function $f(x)=\sqrt{1-g^{2}(x)}$ is defined for those $x$ for which $g(x)$ takes values from the interval [1/4;1]. The set of values of the function $f(x)$ is the set $E_{f}=\lfloor 0 ; \sqrt{15} / 4]$.

Answer: $E_{f}=\lfloor 0 ; \sqrt{15} / 4\rfloor$.