Can you simplify this inverse trig problem?

[Kishlay]

Inverse trig problem -- please help!

Homework Statement

tanx+tan2x+root3tanxtan2x=root3
find x...

Homework Equations

The Attempt at a Solution

i have tried a lot and always ended with a complicated cubic equation...
please help me by giving me a another approach to the solution

 

[jedishrfu]

I'd try using the tan(a+b) identity to reduce the equation into terms with tan(x) factors.

what is the root3 you mention? Is it some arbitrary constant?

 

[Kishlay]

yes I have tried the same thing as you have said
root3 is same as √3

 

[adjacent]

Homework Statement

tanx+tan2x+root3tanxtan2x=root3
find x...

Do you mean:
$tan(x)+tan(2x)+\sqrt{3}*tan(x)*tan(2x)=\sqrt{3}$ ?

 

[jedishrfu]

yes I have tried the same thing as you have said
root3 is same as √3

and you tried using the identity tan(2x) = 2tan(x) / (1-tan^2(x))

and then collecting terms to create a cubic equation in tan(x) that's equal to zero.

 

[Kishlay]

so??

 

[Kishlay]

Do you mean:
$tan(x)+tan(2x)+\sqrt{3}*tan(x)*tan(2x)=\sqrt{3}$ ?

yes..
how did you typed it??

 

[Mark44]

Homework Statement

tanx+tan2x+root3tanxtan2x=root3
find x...

Homework Equations

The Attempt at a Solution

i have tried a lot and always ended with a complicated cubic equation...
please help me by giving me a another approach to the solution

It is not enough to say what you have tried. You need to show us what you have tried.

 

[Kishlay]

Do you mean:
$tan(x)+tan(2x)+\sqrt{3}*tan(x)*tan(2x)=\sqrt{3}$ ?

yes... how did you typed it?

 

[Kishlay]

It is not enough to say what you have tried. You need to show us what you have tried.

i have used the identity of tan(A+B)
then i simplified it and got a cubic equation in tanx
and i don't know how to solve a cubic...
i am a high school student :)

 

[jedishrfu]

yes... how did you typed it?

The PF website provides an advanced editor where you can enter mathematical expressions with subscripts, superscripts, summations, integrals...

Use the "Go Advanced" button to the right of the "Post Quick Reply" button next time.

 

[adjacent]

yes..
how did you typed it??

Latex-Quote it and see how I've written it.You have to write

## and ##

around the sentence.

 

[Kishlay]

Latex-Quote it and see how I've written it.You have to write

## and ##

around the sentence.

ok...:)

 

[jedishrfu]

i have used the identity of tan(A+B)
then i simplified it and got a cubic equation in tanx
and i don't know how to solve a cubic...
i am a high school student :)

So show us the cubic equation.

The goal of the PF forum is help you with your homework not do it for you.

 

[ehild]

No need to solve a cubic equation.
Notice that √3=tan(π/3)

You can rearrange the equation as

$$\tan(2x)(1+\sqrt3 \tan(x))=\sqrt3 - \tan(x)$$

that is

$$\tan(2x)=\frac{\sqrt3 - \tan(x)}{1+\sqrt3 \tan(x)}$$

Recall that

$$\tan(a-b)=\frac{\tan(a) - \tan(b)}{1+\tan(a) \tan(b)}$$



ehild

 

[Kishlay]

It is not enough to say what you have tried. You need to show us what you have tried.

i have got this equation
(tanx)³-3$\sqrt{}3$(tanx)²-3tanx+$\sqrt{}3$ =0

 

[Kishlay]

No need to solve a cubic equation.
Notice that √3=tan(π/3)

You can rearrange the equation as

$$\tan(2x)(1+\sqrt3 \tan(x))=\sqrt3 - \tan(x)$$

that is

$$\tan(2x)=\frac{\sqrt3 - \tan(x)}{1+\sqrt3 \tan(x)}$$

Recall that

$$\tan(a-b)=\frac{\tan(a) - \tan(b)}{1+\tan(a) \tan(b)}$$



ehild

thanks...!

 

[ehild]

You are welcome


ehild

 

[jedishrfu]

dont forget to thank ehild via the thanks button.

 

[ehild]

It had been done and very properly

ehild