Solve for x using trigonometric ratios

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Discussion Overview

The discussion revolves around solving the equation of the form tan(A.x)/tan(B.x)=C for the variable x, utilizing trigonometric ratios and identities. Participants explore various methods, including the addition theorem for sine and numerical solutions.

Discussion Character

  • Mathematical reasoning, Technical explanation, Homework-related

Main Points Raised

  • One participant presents the equation tan(A.x)/tan(B.x)=C and seeks a method to solve for x.
  • Another participant suggests using the addition theorem for sine, proposing the equation ##\sin(Ax) \cos(Bx) = C \sin(Bx) \cos(Cx)## as a potential approach.
  • A similar suggestion is reiterated by another participant, emphasizing the use of the addition theorem and numerical methods, such as WolframAlpha.
  • One participant expresses confusion about the disappearance of the constant C in their manipulation of the equation and suggests using the property ##-\sin(\alpha)=\sin(-\alpha)## to apply the arcsine function.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method to solve for x, and multiple approaches are presented without resolution.

Contextual Notes

There are unresolved steps in the mathematical manipulation, particularly regarding the treatment of the constant C and the application of trigonometric identities.

Sourav Guha
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I have a mathematical term of the form tan(A.x)/tan(B.x)=C.How do I find out the value of x?
 
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I would try the addition theorem for sine, since you have ##\sin(Ax) \cos(Bx) = C \sin(Bx) \cos(Cx)##. Or solve it numerically, e.g. by the help of WolframAlpha.com.
 
fresh_42 said:
I would try the addition theorem for sine, since you have ##\sin(Ax) \cos(Bx) = C \sin(Bx) \cos(Cx)##. Or solve it numerically, e.g. by the help of WolframAlpha.com.
By using Addition Theorem,I finally get sin(A+B)x=-sin(A-B)x..Now how do I find x?
 
Sourav Guha said:
By using Addition Theorem,I finally get sin(A+B)x=-sin(A-B)x..Now how do I find x?
I wonder where the ##C## has gone. But if you have what you say, then we can use ##-\sin(\alpha)=\sin(-\alpha)## and then you can apply ##\arcsin##.
 

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