Solving Goniometric Equation: sen(2x) * sen(x)=sen(4x)*sen(3x)

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Homework Help Overview

The discussion revolves around solving the goniometric equation sen(2x) * sen(x) = sen(4x) * sen(3x), focusing on the transformation of the equation using trigonometric identities.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss applying product to sum and sum to product identities, leading to the equation cos(3x) = cos(7x). Questions arise regarding the implications of this equation and the nature of its solutions.

Discussion Status

Participants are exploring various approaches to solve the transformed equation, including the use of graphical representations and identities. There is a focus on understanding the relationship between angles when cosine values are equal, though no consensus on the solutions has been reached.

Contextual Notes

Some participants express uncertainty about how to proceed with finding all solutions, indicating a need for further clarification on the implications of the cosine equation and the graphical approach.

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Homework Statement


sen(2x) * sen(x) = sen(4x) * sen(3x)

The Attempt at a Solution


I applied product to sum and sum to product identities and now I get cos(3x)=cos(7x ) how can I solve it?

thank you
 
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hi scientifico! :smile:
scientifico said:
I applied product to sum and sum to product identities and now I get cos(3x)=cos(7x)

yup! :smile:

so 3x = ±7x + … ? :wink:

(alternatively, use the formula for cosA - cosB)
 
tiny-tim said:
so 3x = ±7x + … ? :wink:
What is this?
 
draw the graph of cosθ …

what has to be the relation between θ1 and θ2 if cosθ1 = cosθ2 ? :wink:
 
θ1 = θ2 ?
 
yes, but what are all the other solutions?

(and have you drawn a graph of cosθ ?)
 
Yes but i don't know what to search for
 
If cos(3x) = cos(7x), then you have cos(7x) - cos(3x) = 0 .

Use the sum to product identities to change this to the product of two sines .

[itex]\displaystyle \cos \theta - \cos \varphi = -2\sin\left( {\theta + \varphi \over 2}\right) \sin\left({\theta - \varphi \over 2}\right)[/itex]

It's almost always easier to solve an equation with product that equals zero than one with a sum/difference that equals zero.
 
ok but that way I get cos(5x)
 
  • #10
no, you get 2sin(5x)sin(2x) = 0 :confused:
 

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