Giải các phương trình bài 2

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Giải các phương trình bài 2
giai-cac-phuong-trinh-bai-2

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Amity 1 year 2020-11-27T07:29:32+00:00 2 Answers 50 views 0

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  1. Edana
    0
    2020-11-27T07:30:42+00:00
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    `a) sin x = 1`

    `<=> x = π/2 + k2π`

    `b) cos 2x = -1`

    `<=> 2x = π + k2π`

    `<=> x = π/2 + kπ`

    `c) tan 3x = 0`

    `<=> 3x = kπ`

    `<=> x = k(π)/2`

    `d) cot 5x = -1`

    `<=> 5x = -π/4 + kπ`

    `<=> x = -π/20 + k(π)/5`

    `e) sin (3x – π/3) = (\sqrt{3})/2`

    `<=>` \(\left[ \begin{array}{l}3x – \dfrac{π}{3} = \dfrac{π}{3} + k2π\\3x – \dfrac{π}{3} = \dfrac{2π}{3} + k2π\end{array} \right.\) 

    `<=>` \(\left[ \begin{array}{l}x = \dfrac{2π}{9} + k\dfrac{2π}{3}\\x = \dfrac{π}{3} + k\dfrac{2π}{3}\end{array} \right.\) `(k ∈ ZZ)`

    `f) cos (2x – π/4) = (\sqrt{2})/2`

    `<=>` \(\left[ \begin{array}{l}2x – \dfrac{π}{4} = \dfrac{π}{4} + k2π\\2x – \dfrac{π}{4} = -\dfrac{π}{4} + k2π\end{array} \right.\) 

    `<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{4} + kπ\\x = kπ\end{array} \right.\) `(k ∈ ZZ)`

    `g) tan ((3x)/2 + π/6) = sqrt{3}`

    `<=> (3x)/2 + π/6 = π/3 + kπ`

    `<=> (3x)/2 = π/6 + kπ`

    `<=> x = π/9 + k(2π)/3` `(k ∈ ZZ)`

  2. Nho
    0
    2020-11-27T07:31:07+00:00
    Reply

    Đáp án:

    $\begin{array}{l}
    1)\sin x = 1\\
     \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \\
    2)\\
    \cos 2x =  – 1\\
     \Leftrightarrow 2x = \pi  + k2\pi \\
     \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
    3)\tan 3x = 0\\
     \Leftrightarrow 3x = k\pi \\
     \Leftrightarrow x = \dfrac{{k\pi }}{3}\\
    4)cot5x =  – 1\\
     \Leftrightarrow 5x =  – \dfrac{\pi }{4} + k\pi \\
     \Leftrightarrow x =  – \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{5}\\
    5)\sin \left( {3x – \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}\\
     \Leftrightarrow \left[ \begin{array}{l}
    3x – \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\
    3x – \dfrac{\pi }{3} = \dfrac{{2\pi }}{3} + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}\\
    x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}
    \end{array} \right.\\
    6)\cos \left( {2x – \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\
     \Leftrightarrow \left[ \begin{array}{l}
    2x – \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
    2x – \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi 
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{4} + k\pi \\
    x = \dfrac{\pi }{2} + k\pi 
    \end{array} \right.\\
    7)\tan \left( {\dfrac{{3x}}{2} + \dfrac{\pi }{6}} \right) = \sqrt 3 \\
     \Leftrightarrow \dfrac{{3x}}{2} + \dfrac{\pi }{6} = \dfrac{\pi }{3} + k\pi \\
     \Leftrightarrow x = \dfrac{\pi }{9} + \dfrac{{k2\pi }}{3}
    \end{array}$

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