Register Now

Login

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

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

Toán Lớp 11: 4sin^2x – 3sinx.cosx – cos^2x = 0

Toán Lớp 11: 4sin^2x – 3sinx.cosx – cos^2x = 0

Comments ( 1 )

  1. Giải đáp:
    \[\left[ \begin{array}{l}
    x = \dfrac{\pi }{4} + k\pi \\
    x = \arctan \dfrac{{ – 1}}{4} + k\pi 
    \end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
    Lời giải và giải thích chi tiết:
    \(\begin{array}{l}
    4{\sin ^2}x – 3\sin x.\cos x – {\cos ^2}x = 0\,\,\,\,\left( 1 \right)\\
    TH1:\,\,\,\cos x = 0\\
    \left( 1 \right) \Leftrightarrow 4{\sin ^2}x = 0\\
     \Leftrightarrow \sin x = 0\\
     \Rightarrow {\sin ^2}x + {\cos ^2}x = 0\,\,\,\,\,\left( L \right)\\
    TH2:\,\,\,\,\cos x \ne 0\\
    \left( 1 \right) \Leftrightarrow \,\,\dfrac{{4{{\sin }^2}x – 3\sin x.\cos x – {{\cos }^2}x}}{{{{\cos }^2}x}} = 0\\
     \Leftrightarrow 4.\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} – 3.\dfrac{{\sin x}}{{\cos x}} – 1 = 0\\
     \Leftrightarrow 4{\tan ^2}x – 3\tan x – 1 = 0\\
     \Leftrightarrow \left( {4{{\tan }^2}x – 4\tan x} \right) + \left( {\tan x – 1} \right) = 0\\
     \Leftrightarrow 4\tan x\left( {\tan x – 1} \right) + \left( {\tan x – 1} \right) = 0\\
     \Leftrightarrow \left( {\tan x – 1} \right)\left( {4\tan x + 1} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \tan x – 1 = 0\\
    4\tan x + 1 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    \tan x = 1\\
    \tan x =  – \dfrac{1}{4}
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{\pi }{4} + k\pi \\
    x = \arctan \dfrac{{ – 1}}{4} + k\pi 
    \end{array} \right.\,\,\,\,\left( {k \in Z} \right)
    \end{array}\)

Leave a reply

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