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Cherry
Cherry

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Asked: Tháng Mười 27, 20202020-10-27T21:52:12+00:00 2020-10-27T21:52:12+00:00In: Môn Toán

Giúp mình phần lượng giác với

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Giúp mình phần lượng giác với
giup-minh-phan-luong-giac-voi

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    1. Gerda

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      Gerda
      2020-10-27T21:54:03+00:00Added an answer on Tháng Mười 27, 2020 at 9:54 chiều

      Đáp án:

       

      Giải thích các bước giải:

       `(4sin3x+2\sqrt{3})(2cosx-1)=0`

      `⇔` \(\left[ \begin{array}{l}4sin3x+2\sqrt{3}=0\\2cosx-1=0\end{array} \right.\) 

      `⇔` \(\left[ \begin{array}{l} \left[ \begin{array}{l}3x=-\dfrac{\pi}{3}+k2\pi\ (k \in \mathbb{Z})\\3x=\pi-(-\dfrac{\pi}{3})+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\\x=±\dfrac{\pi}{3}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)

      `⇔` \(\left[ \begin{array}{l} \left[ \begin{array}{l}x=-\dfrac{\pi}{9}+k\dfrac{2\pi}{3}\ (k \in \mathbb{Z})\\x=\dfrac{4\pi}{9}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\\x=±\dfrac{\pi}{3}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)

      Vậy ………..

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    2. Bơ

      Bơ

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      Bo
      2020-10-27T21:54:05+00:00Added an answer on Tháng Mười 27, 2020 at 9:54 chiều

      `(4sin 3x + 2\sqrt{3})(2cos x – 1) = 0`

      `<=>` \(\left[ \begin{array}{l}4sin 3x + 2\sqrt{3} = 0\\2cos x – 1 = 0\end{array} \right.\) 

      `<=>` \(\left[ \begin{array}{l}sin 3x = -\dfrac{\sqrt{3}}{2}\\cos x = \dfrac{1}{2}\end{array} \right.\) 

      `<=>` \(\left[ \begin{array}{l}3x = -\dfrac{π}{3} + k2π\\3x = \dfrac{4π}{3} + k2π\\x = ±\dfrac{π}{3} + k2π\end{array} \right.\) 

      `<=>` \(\left[ \begin{array}{l}x = -\dfrac{π}{9} + k\dfrac{2π}{3}\\x = \dfrac{4π}{9} + k\dfrac{2π}{3}\\x = ± \dfrac{π}{3} + k2π\end{array} \right.\) `(k ∈ ZZ)`

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