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

Toán Lớp 11: Giải phương trình lượng giác cos(x-pi/4)=sin(2x+pi/2)

Toán Lớp 11: Giải phương trình lượng giác cos(x-pi/4)=sin(2x+pi/2)

Comments ( 2 )

  1. Giải đáp:
    $S=\left\{-\dfrac{\pi}{4}+k2\pi;\dfrac{\pi}{12}+k\dfrac{2\pi}{3}\,\bigg{|}\,k\in\mathbb Z\right\}$
    Lời giải và giải thích chi tiết:
    $\cos\left(x-\dfrac{\pi}{4}\right)=\sin\left(2x+\dfrac{\pi}{2}\right)$
    $⇔\cos\left(x-\dfrac{\pi}{4}\right)=\cos\left(-2x\right)$
    $⇔\cos2x=\cos\left(x-\dfrac{\pi}{4}\right)$
    $⇔\left[ \begin{array}{l}2x=x-\dfrac{\pi}{4}+k2\pi\\2x=\dfrac{\pi}{4}-x+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
    $⇔\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k2\pi\\3x=\dfrac{\pi}{4}+k2\pi\end{array} \right.\,\,(k\in\mathbb Z)$
    $⇔\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{12}+k\dfrac{2\pi}{3}\end{array} \right.\,\,(k\in\mathbb Z)$
    Vậy $S=\left\{-\dfrac{\pi}{4}+k2\pi;\dfrac{\pi}{12}+k\dfrac{2\pi}{3}\,\bigg{|}\,k\in\mathbb Z\right\}$.

  2. Giải đáp:
     \(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{12}+\dfrac{k2\pi}{3}\end{array} \right.\) (k inZZ)
    Lời giải và giải thích chi tiết:
    cos(x-pi/4)=sin(2x+pi/2)
    <=>cos(x-pi/4)=cos2x
    <=>\(\left[ \begin{array}{l}2x=x-\dfrac{\pi}{4}+k2\pi\\2x=-(x-\dfrac{\pi}{4})+k2\pi\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k2\pi\\3x=\dfrac{\pi}{4}+k2\pi\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{12}+\dfrac{k2\pi}{3}\end{array} \right.\) (k inZZ)

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