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

Toán Lớp 11: $tan^22x – tan2x.tan6x =2$ Giải pt giúp mình

Toán Lớp 11: $tan^22x – tan2x.tan6x =2$
Giải pt giúp mình

Comments ( 1 )

  1. Ta có $\tan 3x = \dfrac{{3\tan x – {{\tan }^3}x}}{{1 – 3{{\tan }^2}x}}$
    Điều kiện: $\left\{ \begin{array}{l} \cos 2x \ne 0\\ \cos 6x \ne 0 \end{array} \right.$
    $\begin{array}{l} {\tan ^2}2x – \tan 2x\tan 6x = 2\\  \Leftrightarrow {\tan ^2}2x – \tan 2x\tan \left( {3.2x} \right) = 2\\  \Leftrightarrow {\tan ^2}2x – \tan 2x.\dfrac{{3\tan 2x – {{\tan }^3}2x}}{{1 – 3{{\tan }^2}2x}} = 2\\  \Leftrightarrow {\tan ^2}2x – \dfrac{{3{{\tan }^2}2x – {{\tan }^4}2x}}{{1 – 3{{\tan }^2}2x}} = 2\\  \Leftrightarrow {\tan ^2}2x – 3{\tan ^4}2x – 3{\tan ^2}2x + {\tan ^4}2x = 2\left( {1 – 3{{\tan }^2}2x} \right)\\  \Leftrightarrow  – 2{\tan ^4}2x – 2{\tan ^2}2x = 2 – 6{\tan ^2}2x\\  \Leftrightarrow  – 2{\tan ^4}2x + 4{\tan ^2}2x – 2 = 0\\  \Leftrightarrow {\tan ^4}2x – 2{\tan ^2}2x + 1 = 0\\  \Leftrightarrow {\left( {{{\tan }^2}2x – 1} \right)^2} = 0\\  \Leftrightarrow {\tan ^2}2x = 1\\  \Leftrightarrow \tan 2x =  \pm 1\\  \Leftrightarrow 2x =  \pm \dfrac{\pi }{4} + k\pi \\  \Leftrightarrow x =  \pm \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\left( {k \in \mathbb Z} \right)(TM) \end{array}$
     

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