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

Toán Lớp 11: Giải phương trình: √2cos2x-√6sin2x=-2

Tháng Mười Hai 4, 2022 Toán học 2 Comments 0 views

Toán Lớp 11: Giải phương trình: √2cos2x-√6sin2x=-2

Comments ( 2 )

  1. bichhhathu
    bichhhathu
    4 Tháng Mười Hai, 2022 at 1:05 chiều
    Reply
    Giải đáp:
    sqrt(2) cos2 x – sqrt(6) sin2 x = -2 $\\$ <=>  2 sqrt(2) (1/2 cos2 x – 1/2 sqrt(3) sin2 x) =-2 $\\$ <=>  2 sqrt(2) [cos(π/3) cos2 x – sin(π/3) sin2 x]=-2 $\\$ <=> 2 sqrt(2) sin(-2 x + π/6) = -2 $\\$ <=> sin(-2 x + π/6) = -1/sqrt(2) 
    <=> \(\left[ \begin{array}{l}
    -2 x + \frac{\pi}{6} = \frac{5\pi}{4}+k2 π \\ 
    -2 x + \frac{\pi}{6} = \frac{7\pi}{4}+k2 π
    \end{array} \right.\)  (với k\in mathbbZ)
    <=> \(\left[ \begin{array}{l}
    -2 x = \frac{13\pi}{12}+k2 π \\ 
    -2 x = \frac{19\pi}{12}+k2 π
    \end{array} \right.\) (với k\in mathbbZ)
    <=> \(\left[ \begin{array}{l}
    x = \frac{-13\pi}{24}-k π \\ 
    x = \frac{-19\pi}{24}-k π
    \end{array} \right.\) (với k\in mathbbZ)
    Vậy S={$\frac{-13\pi}{24}-k π; \frac{-19\pi}{24}-k π | k \in \mathbb{Z}$}

  2. catlantuong
    catlantuong
    4 Tháng Mười Hai, 2022 at 1:04 chiều
    Reply
    ~rai~
    \(\sqrt{2}\cos2x-\sqrt{6}\sin2x=-2\\\Leftrightarrow \dfrac{1}{2}\cos2x-\dfrac{\sqrt{3}}{2}\sin2x=-\dfrac{\sqrt{2}}{2}\\\Leftrightarrow \cos\left(2x+\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{2}}{2}\\\Leftrightarrow \left[\begin{array}{I}2x+\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\\2x+\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}2x=\dfrac{5\pi}{12}\\2x=-\dfrac{13\pi}{12}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{5\pi}{24}+k\pi\\x=-\dfrac{13\pi}{24}+k\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{\dfrac{5\pi}{24}+k\pi;-\dfrac{13\pi}{24}+k\pi\Big|k\in\mathbb{Z}\right\}.\)

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

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