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

Toán Lớp 11: cos3x+sin2x=0 sin(2x-pi/3)+sin(pi/6+x)=0

Tháng Mười Hai 10, 2022 Toán học 2 Comments 0 views

Toán Lớp 11: cos3x+sin2x=0
sin(2x-pi/3)+sin(pi/6+x)=0

Comments ( 2 )

  1. minhtuquynh
    minhtuquynh
    10 Tháng Mười Hai, 2022 at 12:46 sáng
    Reply
    ~rai~
    \(a)\cos3x+\sin2x=0\\\Leftrightarrow \cos3x=-\sin2x\\\Leftrightarrow \cos3x=\sin(-2x)\\\Leftrightarrow \cos3x=\cos\left(\dfrac{\pi}{2}+2x\right)\\\Leftrightarrow \left[\begin{array}{I}3x=\dfrac{\pi}{2}+2x+k2\pi\\3x=-\dfrac{\pi}{2}-2x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{2}+k2\pi\\5x=-\dfrac{\pi}{2}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{10}+k\dfrac{2\pi}{5}\end{array}\right.\\\text{Vậy S=}\left\{\dfrac{\pi}{2}+k2\pi;-\dfrac{\pi}{10}+k\dfrac{2\pi}{5}\right\}.\\b)\sin\left(2x-\dfrac{\pi}{3}\right)+\sin\left(\dfrac{\pi}{6}+x\right)=0\\\Leftrightarrow \sin\left(x+\dfrac{\pi}{6}\right)=-\sin\left(2x-\dfrac{\pi}{3}\right)\\\Leftrightarrow \sin\left(x+\dfrac{\pi}{6}\right)=\sin\left(\dfrac{\pi}{3}-2x\right)\\\Leftrightarrow \left[\begin{array}{I}x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-2x+k2\pi\\x+\dfrac{\pi}{6}=\pi-\dfrac{\pi}{3}+2x+k2\pi.\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}3x=\dfrac{\pi}{6}+k2\pi\\-x=\dfrac{\pi}{2}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{18}+k\dfrac{2\pi}{3}\\x=-\dfrac{\pi}{2}+k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{\dfrac{\pi}{18}+k\dfrac{2\pi}{3};-\dfrac{\pi}{2}+k2\pi\Big|k\in\mathbb{Z}\right\}.\)
     

  2. thaolanphuobg
    thaolanphuobg
    10 Tháng Mười Hai, 2022 at 12:46 sáng
    Reply
    Giải đáp:
    cos3 x + sin2 x = 0
    <=> cos3 x = – sin2 x
    <=> cos(3 x) = cos(-2 x – π/2)
    <=> \(\left[ \begin{array}{l}
    3x = \frac{3 π}{2} – 2 x + 2πn
    \\
    3x = \frac{-3 π}{2} + 2 x + 2 π n
    \end{array} \right.\) (với n\inZ)
    <=> \(\left[ \begin{array}{l}
    5x = \frac{3 π}{2} + 2πn
    \\
    x = \frac{-3 π}{2} + 2 π n
    \end{array} \right.\) (với n\inZ)
    <=>  \(\left[ \begin{array}{l}
    x = \frac{3 π}{10} + \frac{2πn}{5}
    \\
    x = \frac{-3 π}{2} + 2 π n
    \end{array} \right.\) (với n\inZ)
    Vậy S={(3 π)/10+(2 π n)/5; – (3 π)/2+2 π n | n \inZ}
     $\\$
    ——————-
    sin(2x-pi/3)+sin(pi/6+x)=0
    <=> sin(x + π/6) = sin(2 x + (2 π)/3)
    <=>  \(\left[ \begin{array}{l}
    x + \frac{π}{6} = \frac{π}{3} – 2 x + 2 π n\\
    x + \frac{π}{6} = \frac{2π}{3} + 2 x + 2 π n
    \end{array} \right.\) (với n\inZ)
    <=>  \(\left[ \begin{array}{l}
    3 x = \frac{π}{6}+2 π n
    \\
    -x = \frac{π}{2}+2 π n
    \end{array} \right.\) (với n\inZ)
    <=>  \(\left[ \begin{array}{l}
    x = \frac{π}{18}+\frac{2 π n}{3}
    \\
    x = \frac{-π}{2}-2 π n
    \end{array} \right.\) (với n\inZ)
    Vậy S={π/18+(2 π n)/3; – π/2-2 π n | n\inZ}

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

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