3. Find the set of values of the function $y=f^{[2019]}(x)$, where $f(x)=\log _{0.5}\left(\frac{\sin x}{\sin x+15}\right)$,

![](https://cdn.mathpix.com/cropped/2024_05_06_d960854a9c9c598cc600g-14.jpg?height=114&width=1102&top_left_y=865&top_left_x=197)
Solution: First, let's find the range of the function \( y_{1}=f(x) \). Let \( t=\sin x, t \in[-1 ; 1] \Rightarrow y_{1}=\log _{0.5}(z), \quad z=\frac{t}{t+15}=1-\frac{15}{t+15} \quad \) (the function \( z(t) \) is increasing). For \( t \in[-1 ; 1] \) we have \( z \in\left[-\frac{1}{14} ; \frac{1}{16}\right] \), but by the properties of logarithms \( z>0 \), hence \( z \in\left(0 ; \frac{1}{16}\right] \). The function \( y_{1}=\log _{0.5} z \) is decreasing, and for \( z \in\left(0 ; \frac{1}{16}\right] \) we have \( y_{1} \in[4 ;+\infty) \). The function \( y_{2}=f(f(x)) \) will take the same values as the function \( y_{2}=f\left(y_{1}\right) \) if \( y_{1} \in[4 ;+\infty) \). Since the function \( t=\sin y_{1} \) for \( y_{1} \in[4 ;+\infty) \) takes all values from the interval \([-1 ; 1]\), repeating the above reasoning, we get that the range of the function \( y_{2}=f(f(x)) \) is the interval \([4 ;+\infty)\). And so on, therefore, the range of the function \( y=f^{[2019]}(x) \) is the interval \([4 ;+\infty)\).

Answer: \( E(y)=[4 ;+\infty) \).