how do you make a general solution for this problem? (in radians)

[abel muroi]

I was given this problem tan⁴ x - 13 tan² x + 36 = 0 and was told for find all solutions (i have to give the general solution)

here is my work...


tan⁴ x - 13 tan² x + 36 = 0
(tan² x - 4) (tan² - 9) = 0
(tan x - 2) (tan x + 2) (tan x - 3) (tan x + 3)

tan x = +- 2

tan x = +- 3

tan ^-1 = +- 63.43 degrees

tan^-1 = +- 71.57 degrees


is this the general solution?

x = 7/20pi + k*1/2pi

 

[stapel]

tan⁴ x - 13 tan² x + 36 = 0...give the general solution

tan⁴ x - 13 tan² x + 36 = 0
(tan² x - 4) (tan² - 9) = 0
(tan x - 2) (tan x + 2) (tan x - 3) (tan x + 3)

What happened to the "=0"?

tan x = +- 2
tan x = +- 3

Where did the "equals" signs come from? (FYI: This is an example of the sort of "magical" steps which can cause serious problems, and are often counted as being wrong.)

tan ^-1 = +- 63.43 degrees
tan^-1 = +- 71.57 degrees

What happened to the arguments of the inverse tangents? Where did "x", the "3", and the "2" go?

is this the general solution?

x = 7/20pi + k*1/2pi

How did you arrive that this expression? Are you supposed to be rounding? If you let k = 0, do you get any of the values listed?

Please be complete. Thank you!

 

[abel muroi]

What happened to the "=0"?


Where did the "equals" signs come from? (FYI: This is an example of the sort of "magical" steps which can cause serious problems, and are often counted as being wrong.)


What happened to the arguments of the inverse tangents? Where did "x", the "3", and the "2" go?


How did you arrive that this expression? Are you supposed to be rounding? If you let k = 0, do you get any of the values listed?

Please be complete. Thank you!

ok i'll re write what i wrote..

tan⁴ x - 13 tan² x + 36 = 0
(tan² x - 4) (tan² - 9) = 0
(tan x - 2) (tan x + 2) (tan x - 3) (tan x + 3) = 0

+- 2

+- 3

ok so now i will take the inverse of tangent for +-2 and +-3. And i got +-63.43 and +-71.57 degrees.


as for the general solution. I don't know how to build one. can you show me how?

 

[ksdhart]

Finding a "general solution" is really just asking you to find the set ofall possible answers. Think about it this way - let's say 15 degrees is a solution. Because all trig functions are periodic, adding one period will produce the same value for x, and so that's also a solution. And the same with adding two lengths of the period, three lengths etc. You have your four solutions, and you know the period of tangent, so put them together and you've got the answer already.

It will look something like this: x = [your answer] + n * [the period of the function], where n is any integer, positive or negative.

 

[abel muroi]

Finding a "general solution" is really just asking you to find the set ofall possible answers. Think about it this way - let's say 15 degrees is a solution. Because all trig functions are periodic, adding one period will produce the same value for x, and so that's also a solution. And the same with adding two lengths of the period, three lengths etc. You have your four solutions, and you know the period of tangent, so put them together and you've got the answer already.

It will look something like this: x = [your answer] + n * [the period of the function], where n is any integer, positive or negative.

ok since the period of a tan graph is 180, that means the general solution will look like this...

x = 63.43 + k*pi

x = -63.43 + k*pi

x = 71.57 + k*pi

x = -71.57 + k*pi


is this correct?





ps if this is correct, then does that mean that every trig equation with tan will always have a period of 180?

 

[Otis]

does that mean that every trig equation with tan will always have a period of 180?

Hi. I'm not sure that I understand what you're thinking.

What is it that are you considering, when asking about the period of an equation?

I'll say that the period of tan(x) is always 180 degrees, regardless of how or where tan(x) may appear in an expression.

Cheers :cool:

 

[abel muroi]

Hi. I'm not sure that I understand what you're thinking.

What is it that are you considering, when asking about the period of an equation?

I'll say that the period of tan(x) is always 180 degrees, regardless of how or where tan(x) may appear in an expression.

Cheers :cool:

but i need to know what the period is so that i can make a general solution.

i know how the general solution looks like..

x = [your answer] + n * [the period of the function],

so how can i find the period then?

 

[Steven G]

ok i'll re write what i wrote..

tan⁴ x - 13 tan² x + 36 = 0
(tan² x - 4) (tan² - 9) = 0
(tan x - 2) (tan x + 2) (tan x - 3) (tan x + 3) = 0

+- 2

+- 3

ok so now i will take the inverse of tangent for +-2 and +-3. And i got +-63.43 and +-71.57 degrees.


as for the general solution. I don't know how to build one. can you show me how?

You really do not like Stapel's advise. Unfortunately (for you) I agree with Stapel and I will write what you meant to write.
tan x = +/- 2 or tan x = +/- 3
arctan (2)=..., arctan (-2)=..., arctan(3)=... and arctan(-3)=...
If you write sloppy mathematics you leave room for errors.

 

[abel muroi]

You really do not like Stapel's advise. Unfortunately (for you) I agree with Stapel and I will write what you meant to write.
tan x = +/- 2 or tan x = +/- 3
arctan (2)=..., arctan (-2)=..., arctan(3)=... and arctan(-3)=...
If you write sloppy mathematics you leave room for errors.

yes sorry i meant to write that (i have a big test coming up so i'm a little bit exhausted because i have been studying all day)

but i have one more question...

did i get the correct general formulas for the problem i posted?


x = 63.43 + k*pi

x = -63.43 + k*pi

x = 71.57 + k*pi

x = -71.57 + k*pi

i tested each of the solutions and they seem to be correct, but i'm just a little bit worried that i have 4 solutions instead of just 1 or 2.


oh and are these general solutions COMPLETE?

 

[Otis]

did i get the correct general formulas for the problem i posted?

x = 63.43 + k*pi

x = -63.43 + k*pi

x = 71.57 + k*pi

x = -71.57 + k*pi

i'm just a little bit worried that i have 4 solutions instead of just 1 or 2.

:idea: If you replace tan(x) with z, in the given equation, you'll have a 4th-degree polynomial. Such a polynomial has four roots, so it's not suprising that four expressions comprise the general solution.

Regarding your solution above, you have reported both degree measure and radian measure. Convert the degree measure that appears in each line to radian measure.

Also, you need to state the definition for symbol k.

Cheers

PS: Are you still thinking that trigonometric equations have periods?

 

[Steven G]

yes sorry i meant to write that (i have a big test coming up so i'm a little bit exhausted because i have been studying all day)

but i have one more question...

did i get the correct general formulas for the problem i posted?


x = 63.43 + k*pi

x = -63.43 + k*pi

x = 71.57 + k*pi

x = -71.57 + k*pi

i tested each of the solutions and they seem to be correct, but i'm just a little bit worried that i have 4 solutions instead of just 1 or 2.


oh and are these general solutions COMPLETE?

Otis is correct about this being a 4th degree eq in tanx and hence might have 4 solutions. Here is my take on it. Whenever you have tanx = +/- a (and a is not 0, actually assume a>0) then you will have 4 solutions. As there are two quadrants where tan x is positive and tanx = a will have a solution in quad I and III. tan x is negative in the other 2 quads (II and IV) so tan x=-a will have solutions in those quadrants as well.

 

[Otis]

Because all trig functions are periodic, adding one period will produce the same value for x

That x looks like a typo.

Adding one period to x will produce the same value as tan(x).