Mọi người giúp em giải nhanh với ạ

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Mọi người giúp em giải nhanh với ạ
moi-nguoi-giup-em-giai-nhanh-voi-a

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Acacia 4 years 2020-10-17T08:15:40+00:00 1 Answers 87 views 0

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  1. Doris
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    2020-10-17T08:16:43+00:00
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    Đáp án:

    $\begin{array}{l}
    a)Khi:\sin 2x = 0\\
     \Rightarrow 2x = k\pi \\
     \Rightarrow x = \dfrac{{k\pi }}{2}\\
    Pt:2{\cos ^2}2x =  – 4\\
     \Rightarrow {\cos ^2}2x =  – 4\left( {ktm} \right)\\
     + Khi:\sin 2x \ne 0 \Rightarrow x \ne \dfrac{{k\pi }}{2}\\
    Pt:2{\cos ^2}2x – 3\sqrt 3 .2\sin 2x.\cos 2x – 4{\sin ^2}2x =  – 4\\
     \Rightarrow \dfrac{{2{{\cos }^2}2x}}{{{{\sin }^2}2x}} – \dfrac{{3\sqrt 3 \cos 2x}}{{\sin 2x}} – 4 = \dfrac{{ – 4}}{{{{\sin }^2}2x}}\\
     \Rightarrow 2{\cot ^2}2x – 3\sqrt 3 \cot 2x – 4 =  – 4.\dfrac{1}{{{{\sin }^2}2x}}\\
     \Rightarrow 2{\cot ^2}2x – 3\sqrt 3 \cot 2x – 4 + 4.\left( {{{\cot }^2}2x + 1} \right) = 0\\
     \Rightarrow 6{\cot ^2}2x – 3\sqrt 3 \cot 2x = 0\\
     \Rightarrow 3\cot 2x\left( {2\cot 2x – \sqrt 3 } \right) = 0\\
     \Rightarrow \left[ \begin{array}{l}
    \cot 2x = 0\\
    \cot 2x = \dfrac{{\sqrt 3 }}{2}
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    2x = {90^0} + k{.180^0}\\
    2x = {49^0} + k{.180^0}
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x = {45^0} + k{.90^0}\\
    x = 24,{5^0} + k{.90^0}
    \end{array} \right.\left( {tmdk} \right)\\
    b){\sin ^2}x + \sin 2x – 2{\cos ^2}x = \dfrac{1}{2}\\
     + Khi:\cos x = 0 \Rightarrow x = \dfrac{\pi }{2} + k\pi \\
     \Rightarrow {\sin ^2}2x = \dfrac{1}{2}\left( {ktm} \right)\\
     + Khi:\cos x \ne 0 \Rightarrow x \ne \dfrac{\pi }{2} + k\pi \\
     \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{2\sin x.\cos x}}{{{{\cos }^2}x}} – 2 = \dfrac{1}{2}.\dfrac{1}{{{{\cos }^2}x}}\\
     \Rightarrow {\tan ^2}x + 2\tan x – 2 = \dfrac{1}{2}\left( {{{\tan }^2}x + 1} \right)\\
     \Rightarrow \dfrac{1}{2}{\tan ^2}x + 2\tan x – \dfrac{5}{2} = 0\\
     \Rightarrow {\tan ^2}x + 4\tan x – 5 = 0\\
     \Rightarrow \left( {\tan x – 1} \right)\left( {\tan x + 5} \right) = 0\\
     \Rightarrow \left[ \begin{array}{l}
    \tan x = 1\\
    \tan x =  – 5
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{4} + k\pi \left( {tm} \right)\\
    x = \arctan \left( { – 5} \right) + k\pi \left( {tm} \right)
    \end{array} \right.
    \end{array}$

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