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

[karush]

$\tiny{\textbf{7.3.a21 Radford HS}}$
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?

 

[skeeter]

for $0^\circ \le x < 360^\circ$

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

$x = \arcsin\left(\dfrac{1-\sqrt{13}}{6}\right) = 205.74^\circ \text{ and } 334.26^\circ$

 

[topsquark]

$\tiny{\textbf{7.3.a21 Radford HS}}$
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

 

[karush]

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.

 

[topsquark]

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

 

[karush]

thank God for the metric system:)