Giải các phương trình sau

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Giải các phương trình sau
giai-cac-phuong-trinh-sau

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Calantha 4 years 2020-12-01T03:01:57+00:00 1 Answers 52 views 0

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  1. Eirian
    0
    2020-12-01T03:03:50+00:00
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    Đáp án:

    ${6)x = k\dfrac{\pi }{2};x = \dfrac{\pi }{3} + k\pi ;x =  – \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)}$

    $7)x = k\pi \left( {k \in Z} \right)$

    Giải thích các bước giải:

    $\begin{array}{l}
    6){\sin ^2}x = {\sin ^2}2x + {\sin ^2}3x\\
     \Leftrightarrow {\sin ^2}2x + {\sin ^2}3x – {\sin ^2}x = 0\\
     \Leftrightarrow {\sin ^2}2x + \left( {\sin 3x – \sin x} \right)\left( {\sin 3x + \sin x} \right) = 0\\
     \Leftrightarrow {\sin ^2}2x + 2\cos 2x\sin x.2\sin 2x\cos x = 0\\
     \Leftrightarrow {\sin ^2}2x + 2\cos 2x.{\sin ^2}2x = 0\\
     \Leftrightarrow {\sin ^2}2x\left( {1 + 2\cos 2x} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    {\sin ^2}2x = 0\\
    1 + 2\cos 2x = 0
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin 2x = 0\\
    \cos 2x = \dfrac{{ – 1}}{2}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    2x = k\pi \\
    2x = \dfrac{{2\pi }}{3} + k2\pi \\
    2x =  – \dfrac{{2\pi }}{3} + k2\pi 
    \end{array} \right.\left( {k \in Z} \right)\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = k\dfrac{\pi }{2}\\
    x = \dfrac{\pi }{3} + k\pi \\
    x =  – \dfrac{\pi }{3} + k\pi 
    \end{array} \right.\left( {k \in Z} \right)
    \end{array}$

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

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