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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: mọi người ơi :sin4x=cos2x làm sao vậy ạ

Toán Lớp 11: mọi người ơi :sin4x=cos2x làm sao vậy ạ

Comments ( 2 )

  1. $\begin{array}{l}
    \sin 4x = \cos 2x\\
     \Leftrightarrow \cos 2x = \cos \left( {\dfrac{\pi }{2} – 4x} \right)\\
     \Leftrightarrow \left[ \begin{array}{l}
    2x = \dfrac{\pi }{2} – 4x + k2\pi \\
    2x = 4x – \dfrac{\pi }{2} + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}\\
    2x = \dfrac{\pi }{2} – k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}\\
    x = \dfrac{\pi }{4} – k\pi 
    \end{array} \right.\left( {k \in \mathbb{Z}} \right)
    \end{array}$
     

  2. Giải đáp + Lời giải và giải thích chi tiết:
    sin4x=cos2x
    <=>cos(4x-pi/2)=cos2x
    <=>\(\left[ \begin{array}{l}4x-\dfrac{\pi}{2}=2x+k2\pi\\4x-\dfrac{\pi}{2}=-2x+k2\pi\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}2x=\dfrac{\pi}{2}+k2\pi\\6x=\dfrac{\pi}{2}+k2\pi\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\end{array} \right.\)(kinZZ)

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222-9+11+12:2*14+14 = ? ( )