Mn giúp mk câu kia với ạk

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Mn giúp mk câu kia với ạk
mn-giup-mk-cau-kia-voi-ak

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RI SƠ 10 months 2020-11-21T04:25:22+00:00 1 Answers 64 views 0

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    0
    2020-11-21T04:26:53+00:00

    Đáp án:

    \[\left[ \begin{array}{l}
    x = \dfrac{\pi }{4} + k\pi \\
    x = \dfrac{\pi }{3} + k\pi 
    \end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\]

    Giải thích các bước giải:

    Ta có:

    \(\begin{array}{l}
    \sin x.\cos y = \dfrac{1}{2}.\left[ {\sin \left( {x + y} \right) + \sin \left( {x – y} \right)} \right]\\
    2\sqrt 2 \sin \left( {x – \dfrac{\pi }{{12}}} \right).\cos x = 1\\
     \Leftrightarrow \sqrt 2 .\left[ {2\sin \left( {x – \dfrac{\pi }{{12}}} \right).\cos x} \right] = 1\\
     \Leftrightarrow \sqrt 2 .\left[ {\sin \left( {x – \dfrac{\pi }{{12}} + x} \right) + \sin \left( {x – \dfrac{\pi }{{12}} – x} \right)} \right] = 1\\
     \Leftrightarrow \sqrt 2 .\left[ {\sin \left( {2x – \dfrac{\pi }{{12}}} \right) + \sin \left( { – \dfrac{\pi }{{12}}} \right)} \right] = 1\\
     \Leftrightarrow \sin \left( {2x – \dfrac{\pi }{{12}}} \right) + \dfrac{{ – \sqrt 6  + \sqrt 2 }}{4} = \dfrac{1}{{\sqrt 2 }}\\
     \Leftrightarrow \sin \left( {2x – \dfrac{\pi }{{12}}} \right) = \dfrac{{\sqrt 6  + \sqrt 2 }}{4}\\
     \Leftrightarrow \left[ \begin{array}{l}
    2x – \dfrac{\pi }{{12}} = \dfrac{{5\pi }}{{12}} + k2\pi \\
    2x – \dfrac{\pi }{{12}} = \dfrac{{7\pi }}{{12}} + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{4} + k\pi \\
    x = \dfrac{\pi }{3} + k\pi 
    \end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
    \end{array}\)

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