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avatar+271 

Find all solutions of the equation in the interval [0, 2π).

 

5 cos^3 x = 5 cos x

 

I'm not sure if I am approaching this problem properly. 

 Oct 26, 2018
 #1
avatar+98044 
+2

5cos^3 (x)   =  5 cos (x)      ...... subtract 5 cos (x)   from both sides

 

5cos^3(x)  - 5 cos (x)    = 0        .....divide through by  5

 

cos^3(x) - cos (x)  =  0      ....factor

 

cos  (x )  ( cos^2 (x)  - 1)   = 0

 

We have two  equations to consider

 

cos (x )  =  0     and this happens at   pi /2 and 3pi/2    in the given interval

 

Also

 

cos^2 ( x)   - 1   = 0       ....add 1 to both sides

 

cos^2 (x )   =  1      take both roots

 

cos (x)  = 1            or       cos (x)  = -1

This happens at               This happens at

x = 0                                     x =      pi   

 

So...the solutions are

 

x = 0 ,   pi/2,   pi,  3pi/2

 

Here's graph to confirm this : https://www.desmos.com/calculator/jx5rslbflo

 

 

cool cool cool

 Oct 26, 2018
 #2
avatar+271 
+1

Thank you so much.  I would have never got that. 

Ruublrr  Oct 26, 2018
 #3
avatar+98044 
0

Yeah.....you probably would have at some point....some of these trig equations can be real "stumpers"  !!!!

 

 

cool cool cool

CPhill  Oct 26, 2018

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