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

Toán Lớp 11: Giải phương trình lượng giác: sin(x^2 – x) = sin( x + π/3)

Toán Lớp 11: Giải phương trình lượng giác: sin(x^2 – x) = sin( x + π/3)

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    $sin(x² – x) = sin(x + \dfrac{π}{3})$
    TH1 $: x² – x = x + \dfrac{π}{3} + k2π$
    $ ⇔ x² – 2x + 1 = 1 + \dfrac{π}{3} + k2π (k ∈ N)$
    $ ⇔ (x – 1)² = 1 + (\dfrac{1}{3} + 2k)π (k ∈ N)$
    $ ⇔ x = 1 ± \sqrt{1 + (\dfrac{1}{3} + 2k)π}(k ∈ N)$
    TH2 $: x² – x = π – (x + \dfrac{π}{3}) + k2π$
    $ ⇔ x² = (\dfrac{1}{3} + k)2π (k ∈ N)$
    $ ⇔ x = ± \sqrt{(\dfrac{1}{3} + k)2π}(k ∈ N)$
     

  2. $\sin(x^2-x)=\sin\left(x+\dfrac{\pi}{3}\right)$
    $\to \left[ \begin{array}{l}x^2-x=x+\dfrac{\pi}{3}+k2\pi \\x^2-x=\pi-x-\dfrac{\pi}{3}+k2\pi \end{array} \right.$
    $\to \left[ \begin{array}{l}x^2-2x+1=1+\dfrac{\pi}{3}+k2\pi\\x^2=\dfrac{2\pi}{3}+k2\pi\end{array} \right.$
    $\to \left[ \begin{array}{l}(x-1)^2=1+\dfrac{\pi}{3}+k2\pi\\x^2=\dfrac{2\pi}{3}+k2\pi\end{array} \right.$
    $\to \left[ \begin{array}{l}x=1\pm\sqrt{1+\dfrac{\pi}{3}+k2\pi} \\x=\pm\sqrt{\dfrac{2\pi}{3}+k2\pi} \end{array} \right.$ ($k\ge 0; k\in\mathbb{Z}$)

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