- #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.

- Thread starter Petkovsky
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- #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

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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

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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.

Last edited:

- #5

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- #6

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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

Last edited:

- #8

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- #9

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

- #10

Dick

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

- #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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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?So the solution is n*pi ?

- #16

Dick

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It's up to you. But new threads attract more attention than old threads.May i post another problem here or should i open a new thread?

- #17

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

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