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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: Giải phương trình sau 1) sin(-2x+50°) =1/3 2) sin(30°-2x)=-1/4

Tháng Hai 17, 2023 Toán học 2 Comments 0 views

Toán Lớp 11: Giải phương trình sau
1) sin(-2x+50°) =1/3
2) sin(30°-2x)=-1/4

Comments ( 2 )

  1. bichquyenlien
    bichquyenlien
    17 Tháng Hai, 2023 at 11:49 chiều
    Reply
    ~rai~
    \(1)\sin\left(-2x+50^\circ\right)=\dfrac{1}{3}\\\Leftrightarrow \sin\left(-2x+\dfrac{5\pi}{18}\right)=\dfrac{1}{3}\\\Leftrightarrow \left[\begin{array}{I}-2x+\dfrac{5\pi}{18}=\arcsin\dfrac{1}{3}+k2\pi\\-2x+\dfrac{5\pi}{18}=\pi-\arcsin\dfrac{1}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}-2x=\arcsin\dfrac{1}{3}-\dfrac{5\pi}{18}+k2\pi\\-2x=-\arcsin\dfrac{1}{3}+\dfrac{13\pi}{18}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{1}{2}\arcsin\dfrac{1}{3}+\dfrac{5\pi}{36}-k\pi\\x=\dfrac{1}{2}\arcsin\dfrac{1}{3}-\dfrac{13\pi}{36}-k\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\2)\sin(30^\circ-2x)=-\dfrac{1}{4}\\\Leftrightarrow \sin\left(\dfrac{\pi}{6}-2x\right)=-\dfrac{1}{4}\\\Leftrightarrow\left[\begin{array}{I}\dfrac{\pi}{6}-2x=\arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\\dfrac{\pi}{6}-2x=\pi-\arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}2x=-\arcsin\left(-\dfrac{1}{4}\right)+\dfrac{\pi}{6}-k2\pi\\2x=\arcsin\left(-\dfrac{1}{4}\right)-\dfrac{5\pi}{6}-k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{1}{2}arcsin\left(-\dfrac{1}{4}\right)+\dfrac{\pi}{12}-k\pi\\x=\dfrac{1}{2}\arcsin\left(-\dfrac{1}{4}\right)-\dfrac{5\pi}{12}-k\pi.\end{array}\right.\quad(k\in\mathbb{Z})\)

  2. tuyetanhthu
    tuyetanhthu
    17 Tháng Hai, 2023 at 11:49 chiều
    Reply
    \(\begin{array}{l}
    1)\quad \sin(-2x + 50^\circ) = \dfrac13\\
    \Leftrightarrow \sin\left(-2x + \dfrac{5\pi}{18}\right) = \dfrac13\\
    \Leftrightarrow \left[\begin{array}{l}-2x + \dfrac{5\pi}{18} = \arcsin\dfrac13 + k2\pi\\-2x + \dfrac{5\pi}{18} = \pi – \arcsin\dfrac13 + k2\pi\end{array}\right.\\
    \Leftrightarrow \left[\begin{array}{l}x =\dfrac{5\pi}{36} – \dfrac12\arcsin\dfrac13 + k\pi\\x = -\dfrac{13\pi}{36} +\dfrac12\arcsin\dfrac13 + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
    2)\quad \sin\left(30^\circ – 2x\right) = -\dfrac14\\
    \Leftrightarrow \sin\left(2x – \dfrac{\pi}{6}\right) = \dfrac14\\
    \Leftrightarrow \left[\begin{array}{l}2x – \dfrac{\pi}{6} = \arcsin\dfrac14 + k2\pi\\2x – \dfrac{\pi}{6} = \pi – \arcsin\dfrac14 + k2\pi\end{array}\right.\\
    \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + \dfrac12\arcsin\dfrac14 + k\pi\\x =\dfrac{7\pi}{6} – \dfrac12\arcsin\dfrac14 + k\pi\end{array}\right.\quad (k\in\Bbb Z)\end{array}\)

     

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222-9+11+12:2*14+14 = ? ( )

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