Cos4x – sin2x = 0 Tan4x = tan2x Cos4x = sin2x Sin^2(3x) = 0 Cot(3x+1) = tan(x – pi/2)

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1. Đáp án:

   $\begin{array}{l}
   a)\cos 4x – \sin 2x = 0\\
    \Rightarrow 1 – 2{\sin ^2}2x – \sin 2x = 0\\
    \Rightarrow 2{\sin ^2}2x + \sin 2x – 1 = 0\\
    \Rightarrow \left( {2\sin 2x – 1} \right)\left( {\sin 2x + 1} \right) = 0\\
    \Rightarrow \left[ \begin{array}{l}
   \sin 2x = \dfrac{1}{2}\\
   \sin 2x =  – 1
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   2x = \dfrac{\pi }{6} + k2\pi \\
   2x = \dfrac{{5\pi }}{6} + k2\pi \\
   2x = \dfrac{{ – \pi }}{2} + k2\pi 
   \end{array} \right. \Rightarrow \left[ \begin{array}{l}
   x = \dfrac{\pi }{{12}} + k\pi \\
   x = \dfrac{{5\pi }}{{12}} + k\pi \\
   x =  – \dfrac{\pi }{4} + k\pi 
   \end{array} \right.\\
   b)\\
   \tan 4x = \tan 2x\\
    \Rightarrow 4x = 2x + k\pi \\
    \Rightarrow 2x = k\pi \\
    \Rightarrow x = \dfrac{{k\pi }}{2}\\
   c)\cos 4x = \sin 2x\\
    \Rightarrow \sin \left( {\dfrac{\pi }{2} – 4x} \right) = \sin 2x\\
    \Rightarrow \left[ \begin{array}{l}
   \dfrac{\pi }{2} – 4x = 2x + k2\pi \\
   \dfrac{\pi }{2} – 4x = \pi  – 2x + k2\pi 
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x = \dfrac{\pi }{{12}} – \dfrac{{k\pi }}{3}\\
   x =  – \dfrac{\pi }{4} – k\pi 
   \end{array} \right.\\
   d){\sin ^2}3x = 0\\
    \Rightarrow \sin 3x = 0\\
    \Rightarrow 3x = k\pi \\
   x = \dfrac{{k\pi }}{3}\\
   e)\cot \left( {3x + 1} \right) = \tan \left( {x – \dfrac{\pi }{2}} \right)\\
    \Rightarrow \tan \left( {\dfrac{\pi }{2} – 3x – 1} \right) = \tan \left( {x – \dfrac{\pi }{2}} \right)\\
    \Rightarrow \dfrac{\pi }{2} – 3x – 1 = x – \dfrac{\pi }{2} + k\pi \\
    \Rightarrow x = \dfrac{\pi }{4} – \dfrac{1}{4} – \dfrac{\pi }{4}
   \end{array}$

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