What Is the Solution for x in the Equation tan((πx)/2) = √(3)/2?

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Homework Help Overview

The discussion revolves around solving the equation tan((πx)/2) = √(3)/2, which falls under the subject area of trigonometric equations.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different approaches to express x in terms of arctan and periodicity of the tangent function. Some participants attempt to derive x directly from the equation, while others express uncertainty about the correctness of their interpretations.

Discussion Status

There are multiple lines of reasoning being explored, with participants offering different expressions for x. Some guidance is provided regarding the periodic nature of the tangent function, but no consensus has been reached on the final form of the solution.

Contextual Notes

Participants note the importance of understanding the periodicity of the tangent function and question the accuracy of certain values related to the tangent function, indicating a need for clarification on definitions and assumptions.

tonyviet
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Homework Statement



tan((π(x))/2) = √(3)/2

Homework Equations


The Attempt at a Solution


tan(x) = √(3)/2
x= π/6 + πn, since Tan has a period of π
This is where I'm stuck
 
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tan(pi*x/2) = sqrt(3)/2
pi*x/2 = arctan(sqrt(3)/2)

x = (2*arctan(sqrt(3)/2))/pi

What is required of you? Find x?
 


Yea Find x
 


tonyviet said:
Yea Find x

Oh ok no problem
 
Last edited:


I also found an x in your name! Funny that.

Well since for [tex]tan(x)=\frac{\sqrt{3}}{2}[/tex]

[tex]x=\frac{\pi}{6}+\pi n[/tex]

Then if we instead have [tex]tan\left(\frac{\pi x}{2}\right)=\frac{\sqrt{3}}{2}[/tex]

We end up with [tex]\frac{\pi x}{2}=\frac{\pi}{6}+\pi n[/tex]

and now solve for x. Simple, no? :smile:

EDIT: [tex]tan(\pi/6)=1/\sqrt{3}[/tex], not [tex]\sqrt{3}/2[/tex]. Since it's not a nice number, it's best to leave it as [tex]arctan(\sqrt{3}/2)[/tex]
 
Last edited:

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