Trigonometry exam problem

[Petkovsky]

I'm preparing for an entrance exam and i have this problem that is puzzling me.

tan(5x) = tan(3x)

I suppose i should start off by:

sin(5x)/cos(5x) = sin(3x)/cos(3x)

but i don't know what method to use.

Should i break everything down to a single x? Please give me a hint how to start.

 

[cristo]

Should i break everything down to a single x? Please give me a hint how to start.

Sounds like a good idea to me.

 

[Integral]

explore this approach:

tan(4x + x) = tan(4x - x)

 

[snipez90]

Use the identity 2sin(a)cos(b) = sin(a+b) + sin(a-b).

*edit* Or, rather, derive an identity for tan(a) - tan(b) using algebra and knowledge of sin addition/subtraction formulas.

 

[Petkovsky]

Ok i solved it and i received something but I'm not sure if it is correct. May i post it here so you can check it if it's not too much trouble?

 

[snipez90]

What do you have? If my calculations are right, there should only be one value of x that satisfies the equation.

 

[Petkovsky]

I ended up with the quadratic equation:

tan^2(4x)*tanx + tanx = 0

tanx(tan^2(4x) + 1) = 0

From here

tanx = 0 // x = 0
and
tan^2(4x) = -1 //Not possible

 

[snipez90]

Actually, sorry zero is not the only value, I made a logic error. Multiples of 2pi should be the answer. Why?

 

[Petkovsky]

Yes because the period of tan(x) function is 2pi.

 

[Dick]

Yes because the period of tan(x) function is 2pi.

Warning. The period of tan(x) is actually pi.

 

[Petkovsky]

So the solution is n*pi ?

 

[Petkovsky]

OK thank you.

 

[snipez90]

Hmm, sorry I meant multiples of pi, I solved it a different way and didn't check my answers thoroughly :-\.

tan(a)-tan(b) = sin(a-b) / cos(a)*cos(b), we arrive at the equation

sin(2x) / cos(3x)*cos(2x) = 0.

sin(2x) = 0 when x = k(pi) and x = (2k+1)*pi/2 for all integers k. But tan(x) is undefined for the latter set of solutions, so x must be a multiple of pi.

 

[Petkovsky]

May i post another problem here or should i open a new thread?

 

[Dick]

So the solution is n*pi ?

Well, if tan(a)=tan(b) then a-b=n*pi for some integer n, because the period of tan is pi and tan assumes every value only once in it's period. Can you apply that to your problem?

 

[Dick]

May i post another problem here or should i open a new thread?

It's up to you. But new threads attract more attention than old threads.

 

[Petkovsky]

Ok tnx for the tip and for all of your help about this problem :)