Giúp mk vs các cậu ơi

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Giúp mk vs các cậu ơi
giup-mk-vs-cac-cau-oi

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Adela 8 tháng 2020-10-15T17:13:04+00:00 1 Answers 71 views 0

Answers ( )

  1. Đáp án:

    ${1)x = \dfrac{\pi }{4} + k\dfrac{\pi }{2};x = \dfrac{\pi }{{42}} + k\dfrac{{2\pi }}{7};x = \dfrac{{5\pi }}{{42}} + k\dfrac{{2\pi }}{7}\left( {k \in Z} \right)}$

    $2)x = k\pi \left( {k \in Z} \right)$

    Giải thích các bước giải:

    $\begin{array}{l}
    1)\sin 5x + \sin 9x + 2{\sin ^2}x – 1 = 0\\
     \Leftrightarrow \sin 5x + \sin 9x – \cos 2x = 0\\
     \Leftrightarrow 2\sin 7x\cos 2x – \cos 2x = 0\\
     \Leftrightarrow \cos 2x\left( {2\sin 7x – 1} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \cos 2x = 0\\
    2\sin 7x – 1 = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \cos 2x = 0\\
    \sin 7x = \dfrac{1}{2}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    2x = \dfrac{\pi }{2} + k\pi \\
    7x = \dfrac{\pi }{6} + k2\pi \\
    7x = \dfrac{{5\pi }}{6} + k2\pi 
    \end{array} \right.\left( {k \in Z} \right)\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\\
    x = \dfrac{\pi }{{42}} + k\dfrac{{2\pi }}{7}\\
    x = \dfrac{{5\pi }}{{42}} + k\dfrac{{2\pi }}{7}
    \end{array} \right.\left( {k \in Z} \right)\\
    2)4\sin 2x – 3\cos 2x = 3\left( {4\sin x – 1} \right)\\
     \Leftrightarrow 4\sin 2x – 3\cos 2x – 12\sin x + 3 = 0\\
     \Leftrightarrow 8\sin x\cos x – 12\sin x + 3\left( {1 – \cos 2x} \right) = 0\\
     \Leftrightarrow 8\sin x\cos x – 12\sin x + 3.2{\sin ^2}x = 0\\
     \Leftrightarrow 2\sin x\left( {4\cos x + 3\sin x – 6} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x = 0\\
    4\cos x + 3\sin x – 6 = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = k\pi \\
    \dfrac{3}{5}\sin x + \dfrac{4}{5}\cos x = \dfrac{6}{5}
    \end{array} \right.\left( {k \in Z} \right)\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = k\pi \\
    \cos \left( {x – \alpha } \right) = \dfrac{6}{5}(vn)\left( {\alpha :\cos \alpha  = \dfrac{4}{5};\sin \alpha  = \dfrac{3}{5}} \right)
    \end{array} \right.\left( {k \in Z} \right)\\
     \Leftrightarrow x = k\pi \left( {k \in Z} \right)
    \end{array}$

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