# In the expression 3tan(3x) = 7tan(7x), what is x?

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I have these two functions, a(x) = 3tan(3x), and b(x) = 7tan(7x).
What I want to do here is equate them so that I can solve for all of the possible values of x. This gives me ...show more

Best Answer

Hi Alain

The smallest (absolute value wise) solution I can't think of is naturally zero,

but I'll be more organized than to merely suggest my intuitive unfounded solutions.

lets look at the function f(x) = 3tan(3x) - 7tan(7x).

clearly the roots of this function is the desired answer, so lets find them:

Since tan(x) has the period of π,

so tan(3x) and so 3tan(3x) have a period of π/3,

and tan(7x) and so 7tan(7x) have a period of π/7,

and therefore f(x) = 3tan(3x) - 7tan(7x) has the period of lcm(π/3 , π/7).

lcm(π/3 , π/7) = |π/3 * π/7| / gcd(π/3 , π/7) = π²/21 / [π * gcd(1/3 , 1/7)] =

= π/21 / gcd(1/3 , 1/7) = π/21 / gcd(1/3-1/7 , 1/7) = π/21 / gcd(4/21 , 1/7) =

= π/21 / gcd(4/21 , 3/21) = π/21 / gcd(4/21-3/21 , 3/21) = π/21 / gcd(1/21 , 3/21) =

= π/21 / gcd(1/21 , 3/21 - 1/21) = π/21 / gcd(1/21 , 2/21) = π/21 / gcd(1/21 , 2/21 - 1/21)=

= π/21 / gcd(1/21 , 1/21) = π/21 / (1/21) = π.

so f(x) too has the period of π.

If you are unfamiliar with the gcd & lcm function see

http://en.wikipedia.org/wiki/Least_commo...

http://en.wikipedia.org/wiki/Greatest_co...

Since f(x) has a period of π,

we can focus only on the roots in a half open segment of π length,

like (-π/2,π/2], and than for each root "r",

we know about an infinite number of roots of the form r+πk

(with k able to take any integer value).

So now that we limited our search range from R to (-π/2,π/2], lets limit it a beat more:

Note that f is an odd function, meaning that for any x we have f(-x) = -f(x),

proof: f(-x) = 3tan(3(-x)) - 7tan(7(-x)) = 3tan(-3x) - 7tan(-7x) =

= -3tan(3x) - (-7tan(7x)) = -(3tan(3x) - 7tan(7x) = -f(x).

(thanks to "tan" being an odd function !!!)

since f(x) is an odd function than if "r" is a root of "f" than so is "-r", since f(-r) = -f(r) = 0.

So we can limit our search for roots to only non-negative values of x,

so the search in (-π/2,π/2] becomes a search in [0,π/2].

Note that searching in the non-positive half,

must be in p-π/2,0] not (-π/2,0] lest it cost us x=-π/2 and so the possible roots x=πk-π/2 .

At this point there really is no way to go around it but to directly solve 3tan(-3x) - 7tan(-7x).

This will be done by converting both tan(3x) and tan(7x) to rational functions of tan(x),

this can be done by the use of the 2 following identities:

http://desmond.imageshack.us/Himg824/sca...

http://desmond.imageshack.us/Himg4/scale...

now we let y=tan(x) to obtain a simple rational function in each side of the equation.

than by multiplying both sides of the equation by the 2 denominators,

to obtain a simple polynom in both sides,

and so we get a simple polynom of which we need to find roots.

I'm skipping this incredibly hard for a human to do by hand calculation,

since it would take a ridicules amount of time (I let my computer do the hard work).

From some reason my computer also used a half angle identity so y=tan(x/2) not y=tan(x)

Eventually we end up with 5y^6 - 130y^5 + 1483y^4 - 4956y^3 + 1483y^2 - 130y + 5 = 0

which my computer solved by "looking" for the root,

as you may know most polynoms of degree 5 or higher doesn't have a closed form roots.

Although the fact that this polynom is a palindromic one,

might be exploitable to convert the problem to a simpler (lower degree) one.

Once we have the roots for y we use y=tan(x/2) <=> x=2arctan(y) to obtain x.

the only value of x obtained this way that is within (0,π/2]

(remember we already accounted for x=0), is x = 0.85996337014179757318.

So lets go backwards:

x = 0 & x = 0.85996337014179757318 are the only roots of f(x) in [0,π/2].

Since f(x) is odd, we know that for every root "r", "-r" is also a root,

so the roots of f(x) in (-π/2,π/2] are:

x = 0, x = 0.85996337014179757318, x = -0.85996337014179757318

Since f(x) has the period of π,

we know that for every root "r" and a whole (integer) number k, r+πk is also a root of f(x).

So to conclude the roots of f(x) in R are 0.85996337014179757318 c + kπ,

with c able to take the value of 0, 1, or -1, and k able to take any whole (integer) value.

Sorry about the "holes" in my proof,\

but it very might have taken me DAYS to write the whole thing (turing bless the pc ^^),

as you will see if you try to use my explanations of the process,

to actually calculate the whole thing by hand for yourself.

(this partial version took me hours :)

Conclusion: if this was a question given to you by someone who knows it's solution,

that person is either have a fantastic solution I couldn't have come up with,

or that person is highly sadistic ^^.

Hope you found my attempt of assistance helpful.

If you have a question, tell me by "adding details",

and I'll try to get back to you before the question expires.

Good luck

mc²

The smallest (absolute value wise) solution I can't think of is naturally zero,

but I'll be more organized than to merely suggest my intuitive unfounded solutions.

lets look at the function f(x) = 3tan(3x) - 7tan(7x).

clearly the roots of this function is the desired answer, so lets find them:

Since tan(x) has the period of π,

so tan(3x) and so 3tan(3x) have a period of π/3,

and tan(7x) and so 7tan(7x) have a period of π/7,

and therefore f(x) = 3tan(3x) - 7tan(7x) has the period of lcm(π/3 , π/7).

lcm(π/3 , π/7) = |π/3 * π/7| / gcd(π/3 , π/7) = π²/21 / [π * gcd(1/3 , 1/7)] =

= π/21 / gcd(1/3 , 1/7) = π/21 / gcd(1/3-1/7 , 1/7) = π/21 / gcd(4/21 , 1/7) =

= π/21 / gcd(4/21 , 3/21) = π/21 / gcd(4/21-3/21 , 3/21) = π/21 / gcd(1/21 , 3/21) =

= π/21 / gcd(1/21 , 3/21 - 1/21) = π/21 / gcd(1/21 , 2/21) = π/21 / gcd(1/21 , 2/21 - 1/21)=

= π/21 / gcd(1/21 , 1/21) = π/21 / (1/21) = π.

so f(x) too has the period of π.

If you are unfamiliar with the gcd & lcm function see

http://en.wikipedia.org/wiki/Least_commo...

http://en.wikipedia.org/wiki/Greatest_co...

Since f(x) has a period of π,

we can focus only on the roots in a half open segment of π length,

like (-π/2,π/2], and than for each root "r",

we know about an infinite number of roots of the form r+πk

(with k able to take any integer value).

So now that we limited our search range from R to (-π/2,π/2], lets limit it a beat more:

Note that f is an odd function, meaning that for any x we have f(-x) = -f(x),

proof: f(-x) = 3tan(3(-x)) - 7tan(7(-x)) = 3tan(-3x) - 7tan(-7x) =

= -3tan(3x) - (-7tan(7x)) = -(3tan(3x) - 7tan(7x) = -f(x).

(thanks to "tan" being an odd function !!!)

since f(x) is an odd function than if "r" is a root of "f" than so is "-r", since f(-r) = -f(r) = 0.

So we can limit our search for roots to only non-negative values of x,

so the search in (-π/2,π/2] becomes a search in [0,π/2].

Note that searching in the non-positive half,

must be in p-π/2,0] not (-π/2,0] lest it cost us x=-π/2 and so the possible roots x=πk-π/2 .

At this point there really is no way to go around it but to directly solve 3tan(-3x) - 7tan(-7x).

This will be done by converting both tan(3x) and tan(7x) to rational functions of tan(x),

this can be done by the use of the 2 following identities:

http://desmond.imageshack.us/Himg824/sca...

http://desmond.imageshack.us/Himg4/scale...

now we let y=tan(x) to obtain a simple rational function in each side of the equation.

than by multiplying both sides of the equation by the 2 denominators,

to obtain a simple polynom in both sides,

and so we get a simple polynom of which we need to find roots.

I'm skipping this incredibly hard for a human to do by hand calculation,

since it would take a ridicules amount of time (I let my computer do the hard work).

From some reason my computer also used a half angle identity so y=tan(x/2) not y=tan(x)

Eventually we end up with 5y^6 - 130y^5 + 1483y^4 - 4956y^3 + 1483y^2 - 130y + 5 = 0

which my computer solved by "looking" for the root,

as you may know most polynoms of degree 5 or higher doesn't have a closed form roots.

Although the fact that this polynom is a palindromic one,

might be exploitable to convert the problem to a simpler (lower degree) one.

Once we have the roots for y we use y=tan(x/2) <=> x=2arctan(y) to obtain x.

the only value of x obtained this way that is within (0,π/2]

(remember we already accounted for x=0), is x = 0.85996337014179757318.

So lets go backwards:

x = 0 & x = 0.85996337014179757318 are the only roots of f(x) in [0,π/2].

Since f(x) is odd, we know that for every root "r", "-r" is also a root,

so the roots of f(x) in (-π/2,π/2] are:

x = 0, x = 0.85996337014179757318, x = -0.85996337014179757318

Since f(x) has the period of π,

we know that for every root "r" and a whole (integer) number k, r+πk is also a root of f(x).

So to conclude the roots of f(x) in R are 0.85996337014179757318 c + kπ,

with c able to take the value of 0, 1, or -1, and k able to take any whole (integer) value.

Sorry about the "holes" in my proof,\

but it very might have taken me DAYS to write the whole thing (turing bless the pc ^^),

as you will see if you try to use my explanations of the process,

to actually calculate the whole thing by hand for yourself.

(this partial version took me hours :)

Conclusion: if this was a question given to you by someone who knows it's solution,

that person is either have a fantastic solution I couldn't have come up with,

or that person is highly sadistic ^^.

Hope you found my attempt of assistance helpful.

If you have a question, tell me by "adding details",

and I'll try to get back to you before the question expires.

Good luck

mc²

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