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

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The discussion revolves around solving the quadratic equation derived from the trigonometric function \(3\sin^2 x - \sin x - 1 = 0\). The equation is rewritten as \(3u^2 - u - 1 = 0\), leading to solutions for \(u\) expressed as \(u = \frac{1 \pm \sqrt{13}}{6}\). The angles \(x\) in degrees are calculated using the arcsine function, yielding values of approximately 50.14°, 129.86°, 205.74°, and 334.26°. Participants express mixed feelings about the use of quadratic equations in trigonometric contexts, with some noting the elegance of the method while others find it unusual. The conversation highlights the importance of precision in angle measurement, particularly in fields like surveying.
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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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