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: 2sin^3x-sinx=2cos^3x-cosx+cos2x

Toán Lớp 11: 2sin^3x-sinx=2cos^3x-cosx+cos2x

Comments ( 2 )

  1. Lời giải và giải thích chi tiết:
     

    toan-lop-11-2sin-3-sin-2cos-3-cos-cos2

  2. $\begin{array}{l}
    2{\sin ^3}x – \sin x = 2{\cos ^3}x – \cos x + \cos 2x\\
     \Leftrightarrow 2\left( {{{\sin }^3}x – {{\cos }^3}x} \right) + \left( {\cos x – \sin x} \right) – \cos 2x = 0\\
     \Leftrightarrow 2\left( {{{\sin }^3}x – {{\cos }^3}x} \right) – \left( {\sin x – \cos x} \right) + {\sin ^2}x – {\cos ^2}x = 0\\
     \Leftrightarrow 2\left( {\sin x – \cos x} \right)\left( {{{\sin }^2}x + \sin x\cos x + {{\cos }^2}x} \right) – \left( {\sin x – \cos x} \right) + \left( {\sin x – \cos x} \right)\left( {\sin x + \cos x} \right) = 0\\
     \Leftrightarrow \left( {\sin x – \cos x} \right)\left( {2{{\sin }^2}x + 2\sin x\cos x + 2{{\cos }^2}x – 1 + \sin x + \cos x} \right) = 0\\
     \Leftrightarrow \left( {\sin x – \cos x} \right)\left( {2 – 1 + \sin 2x + \sin x + \cos x} \right) = 0\\
     \Leftrightarrow \left( {\sin x – \cos x} \right)\left( {\sin x + \cos x + \sin 2x + 1} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sin x – \cos x = 0\left( 1 \right)\\
    \sin x + \cos x + \sin 2x + 1 = 0\left( 2 \right)
    \end{array} \right.\\
    \left( 1 \right) \Rightarrow \sqrt 2 \sin \left( {x – \dfrac{\pi }{4}} \right) = 0\\
     \Leftrightarrow x – \dfrac{\pi }{4} = k\pi  \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \\
    \left( 2 \right):t = \sin x + \cos x = \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right)\left( {\left| t \right| \le \sqrt 2 } \right)\\
     \Rightarrow \sin 2x = {t^2} – 1\\
    \left( 2 \right) \Leftrightarrow t + {t^2} – 1 + 1 = 0 \Leftrightarrow {t^2} + t = 0\\
     \Leftrightarrow t\left( {t + 1} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    t = 0\\
    t =  – 1
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\
    \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) =  – 1
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x + \dfrac{\pi }{4} = k\pi \\
    x + \dfrac{\pi }{4} =  – \dfrac{\pi }{4} + k2\pi \\
    x + \dfrac{\pi }{4} = \dfrac{{5\pi }}{4} + k2\pi 
    \end{array} \right.\\
     \Leftrightarrow \left[ \begin{array}{l}
    x =  – \dfrac{\pi }{4} + k\pi \\
    x =  – \dfrac{\pi }{2} + k2\pi \\
    x = \pi  + k2\pi 
    \end{array} \right.\left( {k \in Z} \right)\\

    \end{array}$

     

Leave a reply

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