giải giúp e từng bước với ạ

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giải giúp e từng bước với ạ
giai-giup-e-tung-buoc-voi-a

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Dulcie 1 year 2020-11-27T08:51:47+00:00 1 Answers 48 views 0

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  1. Gerda
    0
    2020-11-27T08:53:14+00:00
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    Đáp án:

    $\begin{array}{l}
    e)\left( {\cos \dfrac{x}{2} – 1} \right)\left( {5\sin 2x + 4} \right) = 0\\
     \Rightarrow \left[ \begin{array}{l}
    \cos \dfrac{x}{2} – 1 = 0\\
    5\sin 2x + 4 = 0
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    \cos \dfrac{x}{2} = 1\\
    \sin 2x =  – \dfrac{4}{5}
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    \dfrac{x}{2} = k2\pi \\
    2x = \arcsin \left( { – \dfrac{4}{5}} \right) + k2\pi \\
    2x = \pi  – \arcsin \left( { – \dfrac{4}{5}} \right) + k2\pi 
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x = k4\pi \\
    x = \dfrac{1}{2}\arcsin \left( { – \dfrac{4}{5}} \right) + k\pi \\
    x = \dfrac{\pi }{2} – \dfrac{1}{2}\arcsin \left( { – \dfrac{4}{5}} \right) + k\pi 
    \end{array} \right.\\
    a)sin\left( {2x – \dfrac{\pi }{4}} \right).\left( {2\sin 3x + 1} \right) = 0\\
     \Rightarrow \left[ \begin{array}{l}
    \sin \left( {2x – \dfrac{\pi }{4}} \right) = 0\\
    \sin 3x =  – \dfrac{1}{2}
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    2x – \dfrac{\pi }{4} = k\pi \\
    3x = \dfrac{{ – \pi }}{6} + k2\pi \\
    3x = \dfrac{{7\pi }}{6} + k2\pi 
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
    x = \dfrac{{ – \pi }}{{18}} + k\pi \\
    x = \dfrac{{7\pi }}{{18}} + k\pi 
    \end{array} \right.
    \end{array}$

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