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

Toán Lớp 11: Giải phương trình sau: sinx + cosx – sin2x +1=0

Toán Lớp 11: Giải phương trình sau: sinx + cosx – sin2x +1=0

Comments ( 2 )

  1. $sinx+cosx-sin2x+1=0$
    $⇔sinx+cosx-2sinx.cosx+1=0$ $(*)$
    Đặt $t=sinx+cosx$ 
    $⇔t^2=sin²x+cos²x+2sinx.cosx=1+2sinx.cosx$
    $⇔2sinx.cosx=t^2-1$
    Phương trình $(*)$ trở thành:
    $-t^2+t+2=0$
    $⇔\left[ \begin{array}{l}t=2\\t=-1\end{array} \right.$
    $⇔\left[ \begin{array}{l}sinx+cosx=2\\sinx+cosx=-1\end{array} \right.$
    $⇔\left[ \begin{array}{l}sin\bigg(x+\dfrac{\pi}{4}\bigg)=\sqrt2(L)\\sin\bigg(x+\dfrac{\pi}{4}\bigg)=\dfrac{-\sqrt2}{2}(TM)\end{array} \right.$
    $⇔\left[ \begin{array}{l}x=\dfrac{-\pi}{2}+k2\pi\\x=\pi+k2\pi\end{array} \right.$ $(k∈\mathbb{Z})$
     

  2. $\begin{array}{l} t = \sin x + \cos x\left( { – \sqrt 2  \le t \le \sqrt 2 } \right)\\  \Rightarrow {t^2} = 1 + 2\sin x\cos x = 1 + \sin 2x \Leftrightarrow \sin 2x = {t^2} – 1\\ \sin x + \cos x – \sin 2x + 1 = 0\\  \Leftrightarrow t – \left( {{t^2} – 1} \right) + 1 = 0\\  \Leftrightarrow  – {t^2} + t + 2 = 0\\  \Leftrightarrow {t^2} – t – 2 = 0 \Leftrightarrow \left[ \begin{array}{l} t = 2(L)\\ t =  – 1 \end{array} \right.\\  \Rightarrow \sin x + \cos x =  – 1 \Leftrightarrow \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) =  – 1\\  \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) =  – \dfrac{{\sqrt 2 }}{2}\\  \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{4} =  – \dfrac{\pi }{4} + k2\pi \\ x + \dfrac{\pi }{4} = \dfrac{{5\pi }}{4} + k2\pi  \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x =  – \dfrac{\pi }{2} + k2\pi \\ x = \pi  + k2\pi  \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$  

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