Giải PT: 4cos(2x)-2cos(4x)-cos(8x)-1=0 TÌm min: y=4sin(x)-4cos^2(x)

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1. Đáp án:

   $\begin{array}{l}
   1)\\
   4\cos 2x – 2\cos 4x – \cos 8x – 1 = 0\\
    \Rightarrow 4\cos 2x – 2\cos 4x – \left( {2{{\cos }^2}4x – 1} \right) – 1 = 0\\
    \Rightarrow 4\cos 2x – 2\cos 4x – 2{\cos ^2}4x = 0\\
    \Rightarrow 2\cos 2x – \cos 4x – {\cos ^2}4x = 0\\
    \Rightarrow 2\cos 2x – \cos 4x\left( {1 + \cos 4x} \right) = 0\\
    \Rightarrow 2\cos 2x – cos4x.2co{s^2}2x = 0\\
    \Rightarrow 2\cos 2x\left( {1 – \cos 4x.\cos 2x} \right) = 0\\
    \Rightarrow \left[ \begin{array}{l}
   \cos 2x = 0\\
   1 – \left( {2{{\cos }^2}2x – 1} \right).\cos 2x = 0
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   2x = \frac{\pi }{2} + k\pi \\
    – 2{\cos ^3}2x + \cos 2x + 1 = 0
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\
   \cos 2x = 1
   \end{array} \right. \Rightarrow \left[ \begin{array}{l}
   x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\
   2x = k2\pi 
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\
   x = k\pi 
   \end{array} \right.\\
   2)\\
   y = 4\sin x – 4{\cos ^2}x\\
    = 4\sin x – 4\left( {1 – {{\sin }^2}x} \right)\\
    = 4{\sin ^2}x + 4\sin x – 4\\
    = {\left( {2\sin x + 1} \right)^2} – 5 \ge  – 5\forall x\\
   Vậy\,GTNN\,y =  – 5
   \end{array}$

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