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

Toán Lớp 11: Giải pt: a. cos3x-cos2x-cosx+1=0 b. cos4x-4cos2x+cosx-6=0

Toán Lớp 11: Giải pt:
a. cos3x-cos2x-cosx+1=0
b. cos4x-4cos2x+cosx-6=0

Comments ( 2 )

  1. a. cos3x-cos2x-cosx+1=0
    <=>4cos^3x-3cosx-2cos^2x+1-cosx+1=0
    <=>4cos^3x-2cos^2x-4cosx+2=0
    <=>$\left[\begin{matrix} \cos x=-1\\ \cos x=1\\\cos x =\dfrac{1}{2}\end{matrix}\right.$
    <=> $\left[\begin{matrix} \sin x=0\\ \cos x=\dfrac{1}{2}\end{matrix}\right.$
    <=> $\left[\begin{matrix} x=k\pi\\ x=±\dfrac{\pi}{3}+k2\pi\end{matrix}\right.$(kinZZ)
    b. cos4x-4cos2x+cosx-6=0
    Ta có: 
    + cos4x=2cos^2 2x-1=2(2cos^2x-1)^2-1
    =2(4cos^4x-4cos^2x+1)-1=8cos^4-8cos^2x+1
    +cos2x=2cos^2x-1
    Đặt t=cosx (|t|leq1)
    Phương trình: 8t^4-8t^2+1-4(2t^2-1)+9t-6=0
    <=>8t^4-16t^2+9t-1=0
    <=>(t-1)(8t^3+8t^2-8t+1)=0
    <=>$\left[\begin{matrix} t=1\\ t=\dfrac{1}{2}\\t=\dfrac{-3+\sqrt{13}}{4}&(tm)\\t=\dfrac{-3-\sqrt{13}}{4}&(loại)&\end{matrix}\right.$
    <=>$\left[\begin{matrix} x=k2\pi\\ x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{3}+k2\pi\\x=±\arccos(\dfrac{-3+\sqrt{13}}{4})+k2\pi&\end{matrix}\right.$(kinZZ)

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