Question

Find approximate solutions for the equation in the interval [-pi,pi]:
tan x = 3 csc x

Answer 1

@Hero

Answer 2

i got to \(cos^2x+3cos~x-1=0\) im stuck after that

Answer 3

\[\frac{\sin x}{\cos x} = \frac{3}{\sin x}, [-\pi, \pi]\]

Answer 4

yes

Answer 5

sin^2x - 3cos x = 0
(1 - cos^2x) - 3cos x = 0

Answer 6

yes and that can be turned into cos^2(x)+3cos(x)-1=0

Answer 7

Okay, so all you have to do is let cos(x) = y

Answer 8

y^2 + 3y - 1 = 0

Answer 9

Then factor

Answer 10

dont you factor?

Answer 11

there we go, that's what i was missing

Answer 12

Your study guide doesn't tell you what to do?

Answer 13

nope, just questions

Answer 14

Well, that's pointless. Why would they not show you?

Answer 15

no the teacher gave a semester exam study guide with only questions for us to practice ...she explained briefly but i forgot what she had done cuz i wasnt writing :P

Answer 16

k so i did quadratic and got\[x={{3\pm\sqrt{5}}\over2}\]

Answer 17

i meant \(x=\large{{{3\pm\sqrt{13}}\over2}}\)

Answer 18

@Hero

Answer 19

@rajathsbhat

Answer 20

@hartnn

Answer 21

You need to replace y with cos(x)

Answer 22

yea i did....and then i did quadratic equation

Answer 23

No, I mean replace it back

Answer 24

\[cos~x={{3\pm\sqrt{5}}\over2}?\]

Answer 25

\[\cos(x) = {{3\pm\sqrt{5}}\over2}\]

Answer 26

\[cos~x={{3\pm\sqrt{13}}\over2}\]

Answer 27

okay then what?

Answer 28

Isolate x

Answer 29

\[x=cos^{-1}\left({{3\pm\sqrt{13}}\over2}\right)?\]

Answer 30

Yes, but one result will work, the other won't

Answer 31

im supposed to get \(\pm1.263\) and i got 1.878 :S

Answer 32

@hartnn input?

Answer 33

i got