4sin3x.sinx+4cos(3x-pi/4).cos(x+pi/4)-cos^2(2x+pi/4)=0

0 bình luận về “4sin3x.sinx+4cos(3x-pi/4).cos(x+pi/4)-cos^2(2x+pi/4)=0”

1. $$\eqalign{
   & 4\sin 3x.\sin x + 4\cos \left( {3x – {\pi \over 4}} \right).\cos \left( {x + {\pi \over 4}} \right) – {\cos ^2}\left( {2x + {\pi \over 4}} \right) = 0 \cr
   & \Leftrightarrow 4.{1 \over 2}\left( {\cos 2x – \cos 4x} \right) + 4.{1 \over 2}\left[ {\cos 4x + \cos \left( {2x – {\pi \over 2}} \right)} \right] – {{1 – \cos \left( {4x + {\pi \over 2}} \right)} \over 2} = 0 \cr
   & \Leftrightarrow 2\left( {\cos 2x – \cos 4x} \right) + 2\left[ {\cos 4x + \cos \left( {{\pi \over 2} – 2x} \right)} \right] – {1 \over 2}\left[ {1 – \cos \left( {4x + {\pi \over 2}} \right)} \right] = 0 \cr
   & \Leftrightarrow 2\cos 2x – 2\cos 4x + 2\cos 4x + 2\sin 2x – {1 \over 2}\left( {1 + \sin 4x} \right) = 0 \cr
   & \Leftrightarrow 4\cos 2x + 4\sin 2x – 1 – \sin 4x = 0 \cr
   & \Leftrightarrow 4\left( {\sin 2x + \cos 2x} \right) = {\sin ^2}2x + {\cos ^2}2x + 2\sin 2x\cos 2x \cr
   & \Leftrightarrow 4\left( {\sin 2x + \cos 2x} \right) = {\left( {\sin 2x + \cos 2x} \right)^2} \cr
   & \Leftrightarrow \left( {\sin 2x + \cos 2x} \right)\left( {\sin 2x + \cos 2x – 4} \right) = 0 \cr
   & \Leftrightarrow \left[ \matrix{
   \sin 2x + \cos 2x = 0 \hfill \cr
   \sin 2x + \cos 2x = 4\,\,\left( {Vo\,\,nghiem\,\,do\,\,{1^1} + {1^2} < {4^2}} \right) \hfill \cr} \right. \cr & \Leftrightarrow \sin 2x = - \cos 2x \cr & \Leftrightarrow \tan 2x = - 1 \Leftrightarrow 2x = - {\pi \over 4} + k\pi \cr & \Leftrightarrow x = - {\pi \over 8} + {{k\pi } \over 2}\,\,\left( {k \in Z} \right) \cr} $$

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