Tìm GTLN, GTNN các hàm số a) y = cos (4x + $\frac{\pi}{4}$) – cos (4x – $\frac{\pi}{4}$) b) y = cos$^{2}$x – 4cosx + 2021

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1. a)$y=cos\bigg(4x+\dfrac{π}{4}\bigg)-cos\bigg(4x-\dfrac{π}{4}\bigg)\\=-2sin4x.sin\dfrac{π}{4}\\=-\sqrt{2}sin4x$

   Ta có $sin4x=[-1;1]$

   $\Rightarrow y=[-\sqrt{2};\sqrt{2}]$

   $\Rightarrow Min_y=-\sqrt{2}$ khi $sin4x=1$

   $\Leftrightarrow \left[\begin{matrix}4x=\dfrac{π}{2}+k2π\\4x=\dfrac{-3π}{2}+k2π\end{matrix}\right.(k\in Z)$

   $\Leftrightarrow \left[\begin{matrix}x=\dfrac{π}{8}+k2π\\x=\dfrac{-3π}{8}+k2π\end{matrix}\right.(k\in Z)$

   $Max_y=\sqrt{2}$ khi $sin4x=-1$

   $\Leftrightarrow \left[\begin{matrix}4x=\dfrac{3π}{2}+k2π\\4x=\dfrac{-π}{2}+k2π\end{matrix}\right.(k\in Z)$

   $\Leftrightarrow \left[\begin{matrix}x=\dfrac{3π}{8}+k2π\\x=\dfrac{-π}{8}+k2π\end{matrix}\right.(k\in Z)$

   b)$y=cos^2x-4cosx+2021\\=(cosx-2)^2+2017$

   Ta có 

   $cosx=[-1;1]\Rightarrow (cosx-2)^2=[1;9]\\\Rightarrow y=[2018;2026]$

   $Min_y=2018$ khi $(cosx-2)^2=1$ hay $cosx=1$

   $\Leftrightarrow x=0+k2π\ (k\in Z)$

   $Max_y=2026$ khi $(cosx-2)^2=9$ hay $cosx=-1$

   $\Leftrightarrow x=π+k2π\ (k\in Z)$

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