How Can You Further Simplify 2cos2x - 2cosx in Trigonometry?

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Homework Help Overview

The discussion revolves around simplifying the expression 2cos(2x) - 2cos(x) in the context of trigonometry. Participants are exploring methods to further simplify this expression and find values of x that satisfy the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial simplification of the expression and question how to find values of x that satisfy the equation 2cos(2x) - 2cos(x) = 0. There is an exploration of periodicity in cosine and suggestions to visualize the functions involved.

Discussion Status

The discussion is ongoing, with various approaches being considered. Some participants have offered hints and suggestions for further exploration, while others are questioning specific values and methods. There is no explicit consensus on the next steps or final outcomes.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. The discussion includes attempts to draw connections between the functions involved and their properties.

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Homework Statement


2cos2x-2cosx...how do you simplify this further?


Homework Equations





The Attempt at a Solution


2(cos2x-cosx)..but i have to find 0=2cos2x-2cosx so this doesn't really help me.
 
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You're trying to find when 2cos(2x)-2cos(x)=0? Start by what you did... you want to find x such that cos(2x)=cos(x). Hint: cosine is periodic
 
would it be pi/2 and 3pi/2?
 
nope...maybe try drawing cos(2x) and cos(x)?
 
\cos(2x)=2\cos^{2}(x)-1, so 2\cos(2x)-2\cos(x)=2\cos^{2}(x)-1\cos(x)-1=0. Factoring yields (2\cos(x)+1)(\cos(x)-1)=0
 

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