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

Toán Lớp 11: Nhờ mn giúp đỡ 1Sin^4x+cos^4x=1/2

Tháng Mười Một 3, 2022 Toán học 2 Comments 0 views

Toán Lớp 11: Nhờ mn giúp đỡ
1Sin^4x+cos^4x=1/2

Comments ( 2 )

  1. lanminhngoc
    lanminhngoc
    3 Tháng Mười Một, 2022 at 2:50 chiều
    Reply
    cos^4x + sin^4x = 1/2
    <=> (1 – sinx^2)^2 + sinx^4=1/2
    <=> 1 – 2 sin^2x + 2 sin^4x = 1/2
    <=> sin^4x-sin^2x+1/4=0
    <=> (sin^2x-1/2)^2=0
    <=> sin^2x=1/2
    <=> [{:( sin x = 1/sqrt2),( sin x = – 1/sqrt2):}:}
    <=> [{:( x = pi/4+k2pi),( x = [3pi]/4+k2pi),(x=[-pi]/4+k2pi),(x=[5pi]/4+k2pi):}:}(kinZZ)
     <=> x=pi/4+[kpi]/2\(kinZZ)
    Vậy x=pi/4+[kpi]/2\(kinZZ)

  2. dananhthumai
    dananhthumai
    3 Tháng Mười Một, 2022 at 2:49 chiều
    Reply
    $sin^4x+cos^4x=\dfrac{1}{2}$
    $⇔(sin^2x)^2+2sin^2xcos^2x+(cos^2x)^2-2sin^2xcos^2x=\dfrac{1}{2}$
    $⇔(sin^2x+cos^2x)^2-2sin^2xcos^2x=\dfrac{1}{2}$
    $⇔1-2sin^2xcos^2x=\dfrac{1}{2}$
    $⇔1-\dfrac{1}{2}(2sinxcosx)^2=\dfrac{1}{2}$
    $⇔1-\dfrac{1}{2}(sin^22x)=\dfrac{1}{2}$
    $⇔sin^22x-1=0$
    $⇔$\(\left[ \begin{array}{l}sin2x=1=sin(\dfrac{\pi}{2})\\sin2x=-1=sin(\dfrac{-\pi}{2})\end{array} \right.\) 
    $⇔$\(\left[ \begin{array}{l}2x=\dfrac{\pi}{2}+k2\pi\\2x=\pi-\dfrac{\pi}{2}+k2\pi\end{array} \right.\) hay \(\left[ \begin{array}{l}2x=-\dfrac{\pi}{2}+k2\pi\\2x=\pi+\dfrac{\pi}{2}+k2\pi\end{array} \right.\) 
    $⇔$\(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{5\pi}{4}+k\pi\end{array} \right.\) hay \(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{7\pi}{4}+k\pi\end{array} \right.\) 

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

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