Giúp mk giải vs .cảm ơn trc ạ ????

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Giúp mk giải vs .cảm ơn trc ạ ????
giup-mk-giai-vs-cam-on-trc-a

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Kiệt Gia 6 years 2020-10-24T09:19:17+00:00 1 Answers 200 views 0

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  1. Acacia
    0
    2020-10-24T09:20:26+00:00
    Reply

    Đáp án:

    4) \(\left[ \begin{array}{l}
    x = \dfrac{\pi }{6} + k2\pi \\
    x = \dfrac{\pi }{{30}} + \dfrac{{k2\pi }}{5}
    \end{array} \right.\)

    Giải thích các bước giải:

    \(\begin{array}{l}
    1)\sin \left( {2x + \dfrac{\pi }{6}} \right) = \sin x\\
     \to \left[ \begin{array}{l}
    2x + \dfrac{\pi }{6} = x + k2\pi \\
    2x + \dfrac{\pi }{6} = \pi  – x + k2\pi 
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    x =  – \dfrac{\pi }{6} + k2\pi \\
    3x = \dfrac{{5\pi }}{6} + k2\pi 
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    x =  – \dfrac{\pi }{6} + k2\pi \\
    x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
    \end{array} \right.\left( {k \in Z} \right)\\
    3)\sin \left( {2x – \dfrac{\pi }{4}} \right) + \sin x = 0\\
     \to 2\sin \left( {\dfrac{{2x – \dfrac{\pi }{4} + x}}{2}} \right).\cos \left( {\dfrac{{2x – \dfrac{\pi }{4} – x}}{2}} \right) = 0\\
     \to \sin \left( {\dfrac{{3x – \dfrac{\pi }{4}}}{2}} \right).\cos \left( {\dfrac{{x – \dfrac{\pi }{4}}}{2}} \right) = 0\\
     \to \left[ \begin{array}{l}
    \sin \left( {\dfrac{{3x}}{2} – \dfrac{\pi }{8}} \right) = 0\\
    \cos \left( {\dfrac{x}{2} – \dfrac{\pi }{8}} \right) = 0
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    \dfrac{{3x}}{2} – \dfrac{\pi }{8} = k\pi \\
    \dfrac{x}{2} – \dfrac{\pi }{8} = \dfrac{\pi }{2} + k\pi 
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    x = \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\\
    x = \dfrac{{5\pi }}{4} + k2\pi 
    \end{array} \right.\left( {k \in Z} \right)\\
    2)\sin \left( {3x – \dfrac{\pi }{4}} \right) = \sin x\\
     \to \left[ \begin{array}{l}
    3x – \dfrac{\pi }{4} = x + k2\pi \\
    3x – \dfrac{\pi }{4} = \pi  – x + k2\pi 
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    2x = \dfrac{\pi }{4} + k2\pi \\
    4x = \dfrac{{5\pi }}{4} + k2\pi 
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    x = \dfrac{\pi }{8} + k\pi \\
    x = \dfrac{{5\pi }}{{16}} + \dfrac{{k\pi }}{2}
    \end{array} \right.\left( {k \in Z} \right)\\
    4)\cos \left( {3x – \dfrac{\pi }{6}} \right) = \cos 2x\\
     \to \left[ \begin{array}{l}
    3x – \dfrac{\pi }{6} = 2x + k2\pi \\
    3x – \dfrac{\pi }{6} =  – 2x + k2\pi 
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    x = \dfrac{\pi }{6} + k2\pi \\
    5x = \dfrac{\pi }{6} + k2\pi 
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    x = \dfrac{\pi }{6} + k2\pi \\
    x = \dfrac{\pi }{{30}} + \dfrac{{k2\pi }}{5}
    \end{array} \right.\left( {k \in Z} \right)
    \end{array}\)

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