Giải các phương trình sau

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

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King 12 months 2020-11-08T00:20:16+00:00 1 Answers 44 views 0

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    2020-11-08T00:22:15+00:00

    Giải thích các bước giải:

    Ta có:

    \(\begin{array}{l}
    15,\\
    \sin 2x = 2\cos x + \sqrt 3 .\left( {\sin x – 1} \right)\\
     \Leftrightarrow 2\sin x.\cos x = 2\cos x + \sqrt 3 \left( {\sin x – 1} \right)\\
     \Leftrightarrow \left( {2\sin x.\cos x – 2\cos x} \right) – \sqrt 3 \left( {\sin x – 1} \right) = 0\\
     \Leftrightarrow 2\cos x.\left( {\sin x – 1} \right) – \sqrt 3 \left( {\sin x – 1} \right) = 0\\
     \Leftrightarrow \left( {\sin x – 1} \right)\left( {2\cos x – \sqrt 3 } \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x = 1\\
    \cos x = \dfrac{{\sqrt 3 }}{2}
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{2} + k2\pi \\
    x =  \pm \dfrac{\pi }{6} + k2\pi 
    \end{array} \right.\\
    16,\\
    2{\sin ^2}\left( {\dfrac{\pi }{4} – 2x} \right) + \sqrt 3 \cos 4x = 4{\cos ^2}x – 1\\
     \Leftrightarrow 2.{\left( {\sin \dfrac{\pi }{4}.\cos 2x – \cos \dfrac{\pi }{4}.\sin 2x} \right)^2} + \sqrt 3 \cos 4x = 4{\cos ^2}x – 1\\
     \Leftrightarrow 2.{\left( {\dfrac{{\sqrt 2 }}{2}\cos 2x – \dfrac{{\sqrt 2 }}{2}\sin 2x} \right)^2} + \sqrt 3 .\cos 4x = 4{\cos ^2}x – 1\\
     \Leftrightarrow 2.{\left( {\dfrac{{\sqrt 2 }}{2}} \right)^2}.{\left( {\cos 2x – \sin 2x} \right)^2} + \sqrt 3 .\cos 4x = 4{\cos ^2}x – 1\\
     \Leftrightarrow 1.\left( {{{\cos }^2}2x – 2.\cos 2x.\sin 2x + {{\sin }^2}2x} \right) + \sqrt 3 \cos 4x = 4{\cos ^2}x – 1\\
     \Leftrightarrow \left( {{{\sin }^2}2x + {{\cos }^2}2x} \right) – 2.\sin 2x.\cos 2x + \sqrt 3 \cos 4x = 4{\cos ^2}x – 1\\
     \Leftrightarrow 1 – \sin 4x + \sqrt 3 .\cos 4x = 4{\cos ^2}x – 1\\
     \Leftrightarrow  – \sin 4x + \sqrt 3 \cos 4x = 2.\left( {2{{\cos }^2}x – 1} \right)\\
     \Leftrightarrow  – \dfrac{1}{2}\sin 4x + \dfrac{{\sqrt 3 }}{2}.\cos 4x = \cos 2x\\
     \Leftrightarrow \cos 4x.\cos \dfrac{\pi }{6} – \sin 4x.\sin \dfrac{\pi }{6} = \cos 2x\\
     \Leftrightarrow \cos \left( {4x + \dfrac{\pi }{6}} \right) = \cos 2x\\
     \Leftrightarrow \left[ \begin{array}{l}
    4x + \dfrac{\pi }{6} = 2x + k2\pi \\
    4x + \dfrac{\pi }{6} =  – 2x + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x =  – \dfrac{\pi }{{12}} + k\pi \\
    x =  – \dfrac{\pi }{{36}} + \dfrac{{k\pi }}{3}
    \end{array} \right.\\
    17,\\
    \sqrt 3 \cos 5x – 2\sin 3x.\cos 2x – \sin x = 0\\
     \Leftrightarrow \sqrt 3 \cos 5x – \left( {\sin \left( {3x + 2x} \right) + \sin \left( {3x – 2x} \right)} \right) – \sin x = 0\\
     \Leftrightarrow \sqrt 3 \cos 5x – \sin 5x – \sin x – \sin x = 0\\
     \Leftrightarrow \dfrac{{\sqrt 3 }}{2}.\cos 5x – \dfrac{1}{2}\sin 5x = \sin x\\
     \Leftrightarrow \cos 5x.\cos \dfrac{\pi }{6} – \sin 5x.\sin \dfrac{\pi }{6} = \sin x\\
     \Leftrightarrow \cos \left( {5x + \dfrac{\pi }{6}} \right) = \cos \left( {\dfrac{\pi }{2} – x} \right)\\
     \Leftrightarrow \left[ \begin{array}{l}
    5x + \dfrac{\pi }{6} = \dfrac{\pi }{2} – x + k2\pi \\
    5x + \dfrac{\pi }{6} = x – \dfrac{\pi }{2} + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{{18}} + \dfrac{{k\pi }}{3}\\
    x =  – \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}
    \end{array} \right.
    \end{array}\)

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