# Trigonometry exam problem

• Petkovsky

#### 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.

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

explore this approach:

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

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.

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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?

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

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

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

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

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

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

So the solution is n*pi ?

OK thank you.

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.

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

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?

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.