- #1

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tan(5x) = tan(3x)

I suppose i should start off by:

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

but i dont know what method to use.

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

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- Thread starter Petkovsky
- Start date

- #1

- 62

- 0

tan(5x) = tan(3x)

I suppose i should start off by:

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

but i dont know what method to use.

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

- #2

cristo

Staff Emeritus

Science Advisor

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

- #3

Integral

Staff Emeritus

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explore this approach:

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

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

- #4

- 1,101

- 3

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.

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

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- #5

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- #6

- 1,101

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- #7

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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

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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- #8

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- #9

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

- #10

Dick

Science Advisor

Homework Helper

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

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

- #11

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So the solution is n*pi ?

- #12

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OK thank you.

- #13

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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.

- #14

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

- #15

Dick

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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?

- #16

Dick

Science Advisor

Homework Helper

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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.

- #17

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

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