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I have to solve this equation within the domain 0 ≤ x < 2π : 3 cos x^2 = 2 cos x + 1

 

I graphed this on desmos as two separate functions. I got the solutions x = {0, 0.81, 2.16, 2.86, 3.27, 3.79, 4.15, 4.50, 4.90, 5.12, 5.55, 5.65, 6.13} but the textbook answers are x = {0, 1.91, 4.37}.

 

What did I do wrong?

 May 10, 2020
 #1
avatar+109519 
+1

The textbook is correct.

I cannot tell you what you did wrong because you have not shown me what you actually graphed.

 

Can you share the address of the desmos graph with us please.

And, if it is not self-evident, explain how you got your answers.

 May 10, 2020
 #2
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+1

Link to the graph: https://www.desmos.com/calculator/krwd98cjlz

 

I looked at the points of intersection and noted the x-value.

Guest May 10, 2020
 #3
avatar+109519 
+1

The problem is with your graph.  You have put the 2 in the wrong place,

 

You have graphed   \(cos(x^2)\)      INSTEAD OF      \((cos\;x)^2\) which is the same as  \(cos^2 x\)

 

Here is what your graph should look like:

 

https://www.desmos.com/calculator/uxk237heoa

 

 

Edit:

You have actually answered the question that you have posted here correctly. I did not notice before.

However, I do not think that you posted the question that you were actually asked. 

Melody  May 10, 2020
edited by Melody  May 10, 2020
 #4
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+1

It's a quadratic in cos(x).

cos(x) = -1/3, or cos(x) = 1.

 May 10, 2020
 #5
avatar+109519 
+1

Gues is also right.

This can be tackled as a quadratic.

 

 

\(3( cos x)^2 = 2 cos x + 1\\ let\;y=cosx\\ 3y^2=2y+1\\ 3y^2-2y-1=0\\ 3y^2-3y+1y-1=0\\ 3y(y-1)+1(y-1)=0\\ (3y+1)(y-1)=0\\ y=-1/3\quad or \quad y=1\\ cosx=-1/3 \quad or \quad cosx=1\\ for\quad 0\le x <2\pi\\ x=\pi\pm acos(1/3)\quad or \quad 0\\ x=\pi\pm 1.231 \quad or \quad 0\\ x=1.910,\;\;\;4.373,\;\;\;0 \)

.
 May 10, 2020

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