6. Find the set of values of the function $f(x)=2 \cos \left(\frac{\pi}{4} \sin (\sqrt{x-3}+2 x+2)\right)$.

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# Solution:

$f(x)=2 \cos \left(\frac{\pi}{4} \sin (\sqrt{x-3}+2 x+2)\right)$ Since $\sqrt{x-3}+2 x+2 \in[8 ; \infty)$, then $\sin (\sqrt{x-3}+2 x+2) \in[-1 ; 1], \quad \frac{\pi}{4} \sin (\sqrt{x-3}+2 x+2) \in[-\pi / 4 ; \pi / 4]$. Therefore, $\cos \left(\frac{\pi}{4} \sin (\sqrt{x-3}+2 x+2)\right) \in[\sqrt{2} / 2 ; 1] \Rightarrow 2 \cos \left(\frac{\pi}{4} \sin (\sqrt{x-3}+2 x+2)\right) \in[\sqrt{2} ; 2]$.

Answer: $[\sqrt{2} ; 2]$.