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

Toán Lớp 11: Giúp e ạ! Giải pt: tan$^{2}$x+sin$^{2}$2x=4cos$^{2}$x

Tháng Hai 2, 2023 Toán học 2 Comments 0 views

Toán Lớp 11: Giúp e ạ!
Giải pt: tan$^{2}$x+sin$^{2}$2x=4cos$^{2}$x

Comments ( 2 )

  1. hothaibinh
    hothaibinh
    2 Tháng Hai, 2023 at 5:03 chiều
    Reply
    ĐK: $\cos x\ne 0\to x\ne \dfrac{\pi}{2}+k\pi$
    $\tan^2x+1+\sin^22x=4\cos^2x+1$
    $\to \dfrac{1}{\cos^2x}+1-\cos^22x-4\cos^2x-1=0$
    $\to \dfrac{1}{\cos^2x}-(2\cos^2x-1)^2-4\cos^2x=0$
    $\to \dfrac{1}{\cos^2x}-4\cos^4x+4\cos^2x-1-4\cos^2x=0$
    $\to \dfrac{1}{\cos^2x}-4\cos^4x-1=0$
    $\to 1-4\cos^6x-\cos^2x=0$
    $\to \cos x=\pm\dfrac{\sqrt2}{2}$
    $\to x=\pm\dfrac{\pi}{4}+k2\pi$ hoặc $x=\pm\dfrac{3\pi}{4}+k2\pi$ 
    Kết hợp, ta có $x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$

  2. ngoclandiep
    ngoclandiep
    2 Tháng Hai, 2023 at 5:02 chiều
    Reply
    Điều kiện xác định $\cos x\ne 0\Rightarrow x\ne \dfrac \pi 2+k\pi$
    $\begin{array}{l} {\tan ^2}x + {\sin ^2}2x = 4{\cos ^2}x\\  \Leftrightarrow \dfrac{1}{{{{\cos }^2}x}} – 1 + {\left( {2\sin x\cos x} \right)^2} = 4{\cos ^2}x\\  \Leftrightarrow \dfrac{1}{{{{\cos }^2}x}} – 1 = 4{\cos ^2}x – 4{\sin ^2}x{\cos ^2}x\\  \Leftrightarrow \dfrac{1}{{{{\cos }^2}x}} – 1 = 4{\cos ^2}x\left( {1 – {{\sin }^2}x} \right)\\  \Leftrightarrow \dfrac{1}{{{{\cos }^2}x}} – 1 = 4{\cos ^4}x\\  \Leftrightarrow 4{\cos ^6}x + {\cos ^2}x – 1 = 0\\  \Leftrightarrow 4{t^3} + t – 1 = 0\left( {t = {{\cos }^2}x} \right)\\  \Leftrightarrow 4{t^3} – 2{t^2} + 2{t^2} – t + 2t – 1 = 0\\  \Leftrightarrow 2{t^2}\left( {2t – 1} \right) + t\left( {2t – 1} \right) + \left( {2t – 1} \right) = 0\\  \Leftrightarrow \left( {2t – 1} \right)\left( {\underbrace {2{t^2} + t + 1}_{ > 0}} \right) = 0\\  \Leftrightarrow t = \dfrac{1}{2}\\  \Leftrightarrow {\cos ^2}x = \dfrac{1}{2}\\  \Leftrightarrow \left[ \begin{array}{l} \cos x = \dfrac{{\sqrt 2 }}{2}\\ \cos x =  – \dfrac{{\sqrt 2 }}{2} \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x =  \pm \dfrac{\pi }{4} + k2\pi \\ x =  \pm \left( {\dfrac{\pi }{2} + \dfrac{\pi }{4}} \right) + k2\pi  \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x =  \pm \dfrac{\pi }{4} + k2\pi \\ x =  \pm \dfrac{{3\pi }}{4} + k2\pi  \end{array} \right.\left( {k \in \mathbb{Z}} \right)\\  \Rightarrow x =  \pm \dfrac{\pi }{4} + k\dfrac{\pi }{2}\left( {k \in \mathbb{Z}} \right)\\ \end{array}$  

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

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