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

Toán Lớp 11: giải hộ em câu này đi ạ sin ²x + 1/2sin2x + sin x =0

Toán Lớp 11: giải hộ em câu này đi ạ sin ²x + 1/2sin2x + sin x =0

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
     sin^2 x+1/2 sin\ 2x+sin\ x=0
    ⇔ sin^2 x+1/2 . 2 sin\ x . cos\ x+sin\ x=0
    ⇔ sin^2 x+sin\ x . cos\ x+sin\ x=0
    ⇔ sin\ x(sin\ x+cos\ x+1)=0
    ⇔ \(\left[ \begin{array}{l}\sin\ x=0\\\sin\ x+cos\ x+1=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}\sin\ x=0\\\sin\ x+cos\ x=-1\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\\sqrt{2}\sin\ (x+\dfrac{\pi}{4})=-1\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\\sin\ (x+\dfrac{\pi}{4})=-\dfrac{1}{\sqrt{2}}\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\\x+\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\x=-\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\\x=\pi+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) 

  2. x ∈ {2*pi*k, 2*pi*k-pi/2, 2*pi*k+pi}, k ∈ Z

    toan-lop-11-giai-ho-em-cau-nay-di-a-sin-1-2sin2-sin-0

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

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