Question

Solve.
tan4x-tan2x=0 interval: (0,pi) for x

Answer 1

Please please please tell me those are exponents.

Answer 2

tan^4(x)-tan^2(x)=0

yes?

Answer 3

No, there's no exponents, sorry

Answer 4

So we gotta apply some of this Double Angle nonsense? Ugh ok here we go...

Answer 5

\[\Large\rm \tan(2\color{orangered}{x})=\frac{2\tan(\color{orangered}{x})}{1-\tan^2(\color{orangered}{x})}\]So do you understand how I'm applying this to our tangent 4x?\[\Large\rm \tan(2\color{orangered}{\cdot2x})=\frac{2\tan(\color{orangered}{2x})}{1-\tan^2(\color{orangered}{2x})}\]

Answer 6

yes

Answer 7

I got this so far....
tan2x(sec^2x)=0 @zepdrix

Answer 8

Hmmm

Answer 9

\[\Large\rm \tan(4x)-\tan(2x)=0\]
\[\Large\rm \frac{2\tan(2x)}{1-\tan^2(2x)}-\tan(2x)=0\]Hmm where did you go from here?
1-tan^2(2x) does not equal sec^2(2x).
Is that what you were trying to do?

Answer 10

I got it from here:\[\tan2x(\frac{ 1+\tan^22x }{ 1-\tan^2x })=0\]

Answer 11

Oh ummm..
I guess that's ok :)
Let's just not forget about that denominator, that's giving us a restriction on our solution. So, yes it's ok to multiply it away, but let's remember that tan^2(2x) cannot equal 1.

Answer 12

I'm not sure why that tan^2(2x) cannot equal to 1...

Answer 13

We have a denominator of \(\Large\rm 1-\tan^2(2x)\).
This is giving us a restriction.

Remember back to those uhhh rational equations in algebra?\[\Large\rm \frac{(x+2)(x-5)}{(x-1)(x+1)}=0\]The denominator always gave us restrictions.
We can not let a zero show up in the denominator, so we find out which values of x make that happen.

Answer 14

Oh I see.

Answer 15

So you have:\[\Large\rm \tan(2x)\sec^2(2x)=0\]Zero Factor property leads us to:\[\Large\rm \tan(2x)=0,\qquad\qquad\qquad \sec^2(2x)=0\]

Answer 16

okay

Answer 17

So tan(2x)=0
2x= 0, pi and 2pi?

Answer 18

so 0 and any multiple of pi? yes.\[\Large\rm 2x=0+k \pi\]Don't forget to divide by 2 to solve for x!

Answer 19

Oh oh, we were given an interval, nice.

Answer 20

\[\Large\rm x=0+\frac{k \pi}{2}\]And then figure out which k values give you an x between 0 and pi.

Answer 21

So x= 0, pi and pi/2

Answer 22

Mmmmmm yaaaaa I think that's correct! :)

Answer 23

Boy that's a tough problem :o

Answer 24

And how about the sec^x=0

Answer 25

\[\Large\rm \sec^2(2x)=0\]So then,\[\Large\rm \frac{1}{\cos^2(2x)}=0\]Again, the bottom is only giving us restrictions.
The numerator tells us when this can be zero.
\(\Large\rm 1=0\).
So clearly never.
No solutions from this factor.
Here is a graph to help convince you :)
https://www.desmos.com/calculator/rmuklzuvzb
Never touches the x-axis ya?

Answer 26

Oh I see. Thank you for explaining that and helping me! Greatly apprectiated! :D

Answer 27

np \c:/