Solving trigonometry equations

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    Trigonometry
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Discussion Overview

The discussion revolves around solving the trigonometric equation cos(x) + 2cos²(x) = 0 within the interval (0, 6π). Participants seek clarification on the solution process and breakdown of the steps involved.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents the equation and expresses confusion about how the solution was derived from the notes.
  • Another participant confirms the approach of factoring the equation as cos(x)(1 + 2cos(x)) = 0 and identifies solutions for cos(x) = 0 and cos(x) = -1/2.
  • Specific solutions are proposed, including x = π/2 and x = 2π/3, with general forms for other solutions based on periodicity.
  • A participant questions whether the interval (0, 6π) corresponds to three full cycles, equating it to 1080 degrees.
  • A humorous correction is made regarding the spelling of "pi," distinguishing it from "pie."

Areas of Agreement / Disagreement

Participants generally agree on the method of factoring the equation and identifying solutions, but there is no consensus on the interpretation of the interval or the completeness of the solution process.

Contextual Notes

Some participants mention specific ranges for n in the general solutions, but these ranges are not uniformly agreed upon. There are also variations in how solutions are expressed, indicating potential confusion or differing interpretations.

fordy2707
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Hi all, can you show me how to calculate this and breakdown how you get to the answer for me to Understand ,I have been shown what is believed to be the answer from the notes but not a clue How it was reached.

Find all the solutions to the following equation at interval 0,6PI

cos(x) + 2cos^2(x)=0

=

Cosx(1+2cosx)=0
Cosx=cos(pie/2)
X=2npie+_pie/2
Cosx=cos(2pie/3)

Many thanks
 
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fordy2707 said:
Hi all, can you show me how to calculate this and breakdown how you get to the answer for me to Understand ,I have been shown what is believed to be the answer from the notes but not a clue How it was reached.

Find all the solutions to the following equation at interval 0,6PI

cos(x) + 2cos^2(x)=0

=

Cosx(1+2cosx)=0
Cosx=cos(pie/2)
X=2npie+_pie/2
Cosx=cos(2pie/3)

Many thanks
your approach is good. 1st you need to find a particular solution then from it all solutions in the given range.

$\cos(x)(1+2\cos(x)) = 0$
gives $\cos(x) = 0$ and we have particular solution $x = \dfrac{\pi}{2}$ givinin solution in the range
$x = \dfrac{\pi}{2}+ n \pi $ ( n is from 0 to 5) and ,$x =- \dfrac{\pi}{2}+ n \pi $ ( n is from 1 to 6) so that value is in the range

2nd set of solution
$\cos(x) = \frac{-1}{2}$ or $ x= \frac{2\pi}{3}$ particular solution

$x = \dfrac{2\pi}{3}+ n \pi $ ( n is from 0 to 5) and ,$x =- \dfrac{2\pi}{3}+ n \pi $ ( n is from 1 to 6) so that value is in the range
 
0-6PI is that the equivalent of 3 full cycles ,1080 degress ?
 
fordy2707 said:
Cosx(1+2cosx)=0
Cosx=cos(pie/2)
X=2npie+_pie/2
Cosx=cos(2pie/3)
(Shake) [math]\pi[/math] is spelled "pi." Pie refers to something you eat for breakfast. (Sun)

-Dan
 
fordy2707 said:
Hi all, can you show me how to calculate this
and breakdown how you get to the answer for me to Understand.
I have been shown what is believed to be the answer from the notes but not a clue how it was reached.

Find all the solutions to the following equation at interval (0,\,6\pi).

. . \cos(x) + 2\cos^2(x) \;=\;0
You have a quadratic equation.

Factor: .\cos x(1 + 2\cos x) \;=\;0

Set each factor equal to zero and solve.

. . \cos x \;=\;0 \quad\Rightarrow\quad x \;=\;\tfrac{\pi}{2} + \pi n

. . 1 + 2\cos x \:=\:0 \quad\Rightarrow\quad \cos x \:=\:-\tfrac{1}{2} \quad\Rightarrow\quad x \:=\:\left( \begin{array}{cc}\frac{2\pi}{3} + 2\pi n \\ \frac{4\pi}{3} + 2\pi n \end{array}\right)


 

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