6. Find the set of values of the function

$$
f(x)=\sqrt{13-g^{2}(x)}, \text { where } g(x)=(13 / 4)-\cos ^{2} x+\sin x \text {. }
$$
# Solution:

$f(x)=\sqrt{13-g^{2}(x)}$, where $g(x)=13 / 4-\cos ^{2} x+\sin x$.

Let's find the range of the function $z=g(x)=13 / 4-\cos ^{2} x+\sin x$. The function $g(x)$ is defined for all real numbers. We will make a substitution. Let $t=\sin x$. Then $z=9 / 4+t^{2}+t=2+(t+1 / 2)^{2}$ for $t \in[-1 ; 1]$, and $E_{g}=[2 ; 17 / 4]$. The function $y=f(x)=\sqrt{13-g^{2}(x)}$ has the same range as the function $y=\sqrt{13-z^{2}}$ for $z \in[2 ; \sqrt{17}]$, since the function $y=\sqrt{13-z^{2}}$ is defined for $z \in[-\sqrt{13} ; \sqrt{13}]$, and $z=g(x)$ takes all values from the interval $\left[2 ; \frac{17}{4}\right]$. The minimum value of the function $y=\sqrt{13-z^{2}}$ on the interval $[2 ; \sqrt{17}]$ is 0, and the maximum value is 3. Therefore, $E_{f}=[0 ; 3]$.

Answer: $E_{f}=[0 ; 3]$.