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이 식에서 x의 해를 구해야 합니다.

x>=0

x=<2pi

 
 Oct 13, 2021
edited by Guest  Oct 13, 2021
edited by Guest  Oct 13, 2021
 #1
avatar+114828 
+1

cos(2x)-5cos(x)+3=0

 

\(cos(2x)-5cos(x)+3=0\\ cos^2x-sin^2x-5cosx+3=0\\ cos^2x-(1-cos^2x)-5cosx+3=0\\ 2cos^2x-1-5cosx+3=0\\ 2cos^2x-5cosx+2=0\\ let\;\;y=cosx\\ 2y^2-5y+2=0\\ y=\frac{5\pm\sqrt{25-16}}{4}\\ y=\frac{5\pm 3}{4}\\ y=\frac{8}{4}\qquad y=\frac{2}{4}\\ cosx=2\qquad or \qquad cosx=0.5\\ cosx=0.5\\ x=2\pi n \pm \frac{\pi}{6}\qquad n\in Z\\ x=\frac{\pi}{6}\quad or \quad x=\frac{11\pi}{6}\qquad 0

 

 

 

 

The LaTex is not rendering.   I have taken a pic of what it is supposed to be looking like.

 


 

 
 Oct 13, 2021
edited by Melody  Oct 13, 2021
edited by Melody  Oct 13, 2021
 #3
avatar+114828 
0

asinus is right.  I did the last bit wrong.

 

x= pi/3      or  x = 5pi/3

 
Melody  Oct 13, 2021
 #2
avatar+12393 
+2

cos(2x)-5cos(x)+3=0

이 식에서 x의 해를 구해야 합니다.

 

Hello Guest!

 

\(cos(2x)=2cos^2(x)-1\\ 2cos^2(x)-1-5cos(x)+3=0\\ cos^2(x)-2.5cos(x)+1=0\\ cos(x)=1.25\pm\sqrt{1.25^2-1}\\ cos(x)\in \{0.5,2(not\ applicable )\}\)

\(x=acos(0.5)\)

\(x\in \{60°,300°\}\)

laugh  !

 
 Oct 13, 2021
edited by asinus  Oct 13, 2021

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