Question

How should I isolate the trig function?

Answer 1

\[2\tan ^4x-\tan ^2x-15=0\]

Answer 2

@jim_thompson5910 Can I factor this to \[(\tan^2x)(2\tan^2-1)=15\]

Answer 3

yeah but it doesn't really help you isolate x

Answer 4

let z = tan^2(x)

so you will have this equation

2z^2 - z - 15 = 0

factor the left side to get

(2z + 5)(z - 3) = 0

solve for z and use the solutions for z to get the solutions for x

Answer 5

woah..never thought of it like that O_O

Answer 6

so 3 and -2.5...what do i plug back in now?

Answer 7

@jim_thompson5910

Answer 8

recall that z = tan^2(x)

the key is that squared, so z has to be nonnegative

Answer 9

3,2.5

Answer 10

so z = -2.5 yields no real solutions for x

Answer 11

z = 3 is the only one that makes sense if you're looking for real numbered solutions

Answer 12

so...do I still plug back in to check somehow...am i suppose to get a an answer in radians?

Answer 13

x is some angle, yes

Answer 14

I don't know if your teacher wants degrees or radians

Answer 15

but beforehand, you mentioned pi and such, so it's probably in radians

Answer 16

actually, going back to my factor on my first post..

tanx^2(2tan^2-1)=0

couldn't I solve for tan here?

Answer 17

The 2nd parenthesis could be tan=1/sqrt 2

Answer 18

you had 15 on the right side, not 0

Answer 19

why not go back to the very first equation given to you

Answer 20

oh, i see now

Answer 21

sometimes when trying to solve for x, solutions are introduced that don't work for the original equation

Answer 22

that's why it's always best to check the ORIGINAL equation

Answer 23

how i plug 3?

Answer 24

\[\Large z = 3\]

\[\Large \tan^2(x) = 3\]

\[\Large \tan(x) = \pm\sqrt{3}\]

\[\Large \tan(x) = \sqrt{3} \text{ or } \tan(x) = -\sqrt{3}\]

\[\Large x = ?? \text{ or } x = ??\]

Answer 25

wait do you have to plug back into 2tan^2-1 or just tan^2x

Answer 26

@jim_thompson5910

Answer 27

what solutions for x did you get?

Answer 28

well before i solve for tan sqrt 3, i was just wondering if i needed to set the other factor which i think is 2tan^2-1...i mean...where did you get tanx^2?

Answer 29

I let \[\Large z = \tan^2(x)\] so I could get the factorization (2z + 5)(z - 3) = 0

Answer 30

oooh

Answer 31

(x=pi/3+npi) and (x=2pi/3+npi)

Answer 32

cause its general solution not a set domain right?

Answer 33

you have the correct solutions for \[\Large \tan(x) = \pm\sqrt{3}\]

Answer 34

like if it under the domain of 0<x<2pi, i would get more solutions?

Answer 35

oh if you have that domain restriction, then you only have 2 solutions

Answer 36

there is no domain restriction but since this is positive and negative, wouldn't there be 4 solutions?

Answer 37

IF there was that domain restriction

Answer 38

sry, yeah 4

Answer 39

correct

Answer 40

things are getting clearer..thank you.

Answer 41

np