Author Topic: Gah, trig again! (Read 4524 times) Tweet Share

/0 · « **on:** April 26, 2008, 07:58:03 pm »

Find the x-intercepts of $y=\sqrt{3}-3\tan{(2x-\frac{\pi}{5})}\ , \ 0 \leq x \leq \pi$

i.e. $\tan{(2x-\frac{\pi}{5})}=\frac{\sqrt{3}}{3}$

Ok this is kinda freaking me out... it's pretty obvious the reference angle here is $\frac{\pi}{6}$ but my calculator can't even find an exact answer!!!

Thanks

Mao · « **Reply #1 on:** April 26, 2008, 08:14:44 pm »

$\tan{(2x-\frac{\pi}{5})}=\frac{\sqrt{3}}{3}$ for $0 \le x \le \pi$

let $u=2x-\frac{\pi}{5}$ , it follows that $-\frac{\pi}{5}\le u \le \frac{9\pi}{5}$

now solve for $\tan(u)=\frac{\sqrt{3}}{3}$ :

$u=\frac{\pi}{6}+n\cdot \pi$ , which yields us 2 solutions within the implied domain of u, $n=\{0,1\}$

the rest you should be able to handle =)

/0 · « **Reply #2 on:** April 26, 2008, 08:24:07 pm »

Thanks Mao!

Chocolate · « **Reply #3 on:** April 27, 2008, 02:34:42 am »

Isn't it just..
$\tan{(2x-\frac{\pi}{5})}=\frac{\sqrt{3}}{3}$

${(2x-\frac{\pi}{5})}= {\frac{\pi}{6}}, {\frac{7\pi}{6}}$

$x= {\frac{11\pi}{60}}, {\frac{41\pi}{60}}$ for domain $[0,\pi]$

I'm missing something, aren't I? lol

/0 · « **Reply #4 on:** April 27, 2008, 06:37:30 am »

For my reference angle I kept thinking it was $-\frac{\sqrt{3}}{3}$ for some reason - that's why i got it wrong... but my calculator really couldn't give an exact answer, weird.

Captain · « **Reply #5 on:** April 27, 2008, 10:20:05 am »

$\tan{(2x-\frac{\pi}{5})}=\frac{\sqrt{3}}{3}$ for $0 \le x \le \pi$

Rearange the domain:

$0 \le 2x \le 2\pi$ Multiplied domain by 2

$0 - \frac{\pi}{5} \le 2x - \frac{\pi}{5} \le 2\pi - \frac{\pi}{5}$ -pi/5

$- \frac{\pi}{5} \le 2x - \frac{\pi}{5} \le \frac{9\pi}{5}$

Now Solve $\tan{(2x-\frac{\pi}{5})}=\frac{\sqrt{3}}{3}$ for $- \frac{\pi}{5} \le 2x - \frac{\pi}{5} \le \frac{9\pi}{5}$

$(2x-\frac{\pi}{5}) = \frac{\pi}{6}, \frac{7\pi}{6}$

$2x = \frac{11\pi}{30}, \frac{41\pi}{30}$

$x = \frac{11\pi}{60}, \frac{41\pi}{60}$

Everyone has there own technique I guess. I love trig

Mao · « **Reply #6 on:** April 27, 2008, 11:18:52 am »

Quote from: DivideBy0 on April 27, 2008, 06:37:30 am

For my reference angle I kept thinking it was $-\frac{\sqrt{3}}{3}$ for some reason - that's why i got it wrong... but my calculator really couldn't give an exact answer, weird.

i'm assuming you are using TI 89:

when you solve:

$\mbox{solve(\tan(2x-\pi /5)=\sqrt{(3)}/3,x)}$

the answer returned is rather generic:

$x=\frac{(30\cdot \mbox{@n_i}-19)\cdot \pi}{60}$

that is, $\mbox{@n_i}\in Z$ : when you substitude an integer into $\mbox{@n_i}$ , it'll be a valid solution.

for this case though, you will need to restrict the domain of x so you can actually come up with an answer:

$\mbox{solve(\tan(2x-\pi /5)=\sqrt{(3)}/3,x)}|0<x<\pi$
or
$\mbox{solve(\tan(2x-\pi /5)=\sqrt{(3)}/3,x)}|0<x \mbox{ and } x<\pi$

and that will give u exact solutions within the right domain.

/0 · « **Reply #7 on:** April 27, 2008, 09:10:30 pm »

Thanks everyone, I've got it now

dcc · « **Reply #8 on:** April 27, 2008, 09:12:33 pm »

calculators are bad for the soul!

learn the hand thing!

Mao · « **Reply #9 on:** April 27, 2008, 09:14:32 pm »

Quote from: dcc on April 27, 2008, 09:12:33 pm

calculators are bad for the soul!

learn the hand thing!

solve by hand:
$x^3+x+1=0$

good luck dcc, and no mathematica!
(and people dont attempt this)

droodles · « **Reply #10 on:** April 27, 2008, 09:21:59 pm »

you shouldn't use ur calc when trying to find exact vales in trig anyway, they won't take decimal answers in exams and SACS.

dcc · « **Reply #11 on:** April 27, 2008, 10:26:11 pm »

$x^3 + x + 1 = 0$

$\mbox{Let } x = y - \dfrac{1}{3y}$

Substituting back in, we get:

$(y - \dfrac{1}{3y})^3 + (y - \dfrac{1}{3y}) + 1 = 0$

expanding:

$y^3 + 3(y^2)(-\dfrac{1}{3y}) + 3(y)(\dfrac{1}{3y})^2 + (-\dfrac{1}{3y})^3 + y - \dfrac{1}{3y} + 1 = 0$

simplifying:

$y^3 - \dfrac{1}{27y^3} + 1 = 0$

multiplying through by y^3:

$y^6 + y^3 - \dfrac{1}{27} = 0$

$\mbox{Let } u = y^3$

$ \begin{align*} 0 &= u^2 + u - \dfrac{1}{27}\\ &= u^2 + u + \dfrac{1}{4} - \dfrac{1}{4} - \dfrac{1}{27}\\ \dfrac{31}{108} &= (u + \dfrac{1}{2})^2\\ \end{align*}$

$u = \pm \sqrt{\dfrac{31}{108}} - \dfrac{1}{2}$ (but im gonna take the negative because cbf solving two, thanks mao)

therefore:

$y^3 = -\sqrt{\dfrac{31}{108}} - \dfrac{1}{2}$

$\implies y = \sqrt[3]{-\sqrt{\dfrac{31}{108}} - \dfrac{1}{2}}$

$\implies x = \sqrt[3]{-\sqrt{\dfrac{31}{108}} - \dfrac{1}{2}} - \dfrac{1}{3(\sqrt[3]{-\sqrt{\dfrac{31}{108}} - \dfrac{1}{2}})}$

$\implies x = \dfrac{\sqrt[3]{(-\sqrt{\dfrac{31}{108}} - \dfrac{1}{2})^2} - \dfrac{1}{3}}{\sqrt[3]{-\sqrt{\dfrac{31}{108}} - \dfrac{1}{2}}}$

Which is as far as I will go.

Exact answer, no mathematica involved

Mao · « **Reply #12 on:** April 27, 2008, 10:27:09 pm »

HOLY SH!T

man, ur a legend

Odette · « **Reply #13 on:** April 27, 2008, 10:29:50 pm »

Impressive Dcc, and you predict 40 for methods... ha! lol

dcc · « **Reply #14 on:** April 27, 2008, 10:32:01 pm »

anyone could do that with the right toolbox

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Author Topic: Gah, trig again! (Read 4524 times) Tweet Share

/0

Gah, trig again!

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