Finding x in Degrees: Solving a Quadratic Equation for Trigonometric Functions

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Discussion Overview

The discussion revolves around solving the quadratic equation \(3\sin^2 x - \sin x - 1 = 0\) for \(x\) in degrees, specifically focusing on the implications of using trigonometric functions and the resulting angle measures.

Discussion Character

  • Mathematical reasoning, Homework-related, Debate/contested

Main Points Raised

  • One participant rewrites the equation as \(3u^2 - u - 1 = 0\) and calculates \(u = \dfrac{1 \pm \sqrt{13}}{6}\), questioning whether the resulting angles will be in decimal degrees.
  • Another participant provides specific angle solutions using the arcsine function, yielding angles of approximately \(50.14^\circ\), \(129.86^\circ\), \(205.74^\circ\), and \(334.26^\circ\).
  • One participant expresses confusion about the necessity of using the quadratic equation, noting that no answer was provided in their reference material.
  • A participant reflects on their experience in surveying, mentioning that angles were typically recorded to four decimal places or in degrees, minutes, and seconds, without the use of trigonometry.
  • Another participant appreciates the elegance of the quadratic method, contrasting it with their experience in physics where answers are often decimal values.
  • A light-hearted comment is made about the metric system.

Areas of Agreement / Disagreement

Participants express varying opinions on the use of the quadratic equation, with some finding it elegant while others are confused by it. There is no consensus on the necessity or appropriateness of the method used.

Contextual Notes

Some participants note the lack of provided answers in their references and the potential for angles to be expressed in different formats, indicating a dependence on context and definitions.

karush
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$\tiny{\textbf{7.3.a21 Radford HS}}$
find x in degrees $\quad 3\sin^2 x -\sin x-1=0$
rewrite as $3u^2-u-1=0$
quadradic eq
$u=\dfrac{1\pm \sqrt{13}}{6}$

ok this is ?? are we going to have decimal degrees?
 
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for $0^\circ \le x < 360^\circ$

$x = \arcsin\left(\dfrac{1+\sqrt{13}}{6}\right) = 50.14^\circ \text{ and } 129.86^\circ$

$x = \arcsin\left(\dfrac{1-\sqrt{13}}{6}\right) = 205.74^\circ \text{ and } 334.26^\circ$
 
karush said:
$\tiny{\textbf{7.3.a21 Radford HS}}$
find x in degrees $\quad 3\sin^2 x -\sin x-1=0$
rewrite as $3u^2-u-1=0$
quadradic eq
$u=\dfrac{1\pm \sqrt{13}}{6}$

ok this is ?? are we going to have decimal degrees?
Unless you just happen to know what [math]asn \left ( \dfrac{1 \pm \sqrt{13}}{6} \right )[/math] is then, yes, your angle will be ugly!

-Dan
 
i just thot having to use quadradic eq was weird
there was no answer given

Well I remember helping in surveying that the degrees were carried out 4 decimal places
or degree minutes secconds
but we never used trig... the instrument did it all for us.
 
karush said:
i just thot having to use quadradic eq was weird
there was no answer given

Well I remember helping in surveying that the degrees were carried out 4 decimal places
or degree minutes secconds
but we never used trig... the instrument did it all for us.
I think the method with the quadratic is rather elegant. And hey, in Physics almost all of the answers are decimals!

-Dan
 
thank God for the metric system:)
 

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