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Solve the following equation, xe[2,0pi]. tan^3x-9tan^2x+17tanx-6=0

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Well I tried putting x=1, I got an error.
\[x \approx3.14159 n+1.10715\] \[x \approx3.14159 n+0.429998\] \[x \approx3.14159 n+1.4191\]
first see if you can solve this \[u^3-9u^2+17u-6=0\]

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Other answers:

Thanks myin; that's the way.
if you try u=2, you will see you have one solution
thats what i wasn gonna say replace tan with a variable
then using synthetic division you can find the other 2 solutions
I dunno what that is.
synethic division+quadratic formula (or perhaps factoring if we have other rational zeros)
u can also urs long division to get factors
Wait a sec guys, I'm trying to understand this all...
A fancy way of saying that as u = 2 is a solution, (u-2) must be a factor of the polynomial. Hence you can write \[ u^3āˆ’9u^2+17^uāˆ’6 = (u-2)q(u) \] for some second order polynomial
make that 17u, not 17^u
What happened to tan?
i'll let the other chefs take over
we will get back to the tan business after we solve that polynomial=0 for u
Ok, so I'm doing the whole F(#) thing at this point right?
what does F(#) mean?
Looking for the value of x that makes everything 0.
u = 2 u = 1/2 (7-sqrt(37)) u = 1/2 (7+sqrt(37))
If you think how myininaya magically found that \( u=2 \) there is a analytical way of doing it which is known as Rational root test.
right fool
What if I got an error on my calculator?
Useless its tanx (as an entirety) that I using....
Your trying to plug number right into the equation?
i went through the possible rational zeros and found a rational zero i didn't go through all possible rational zeros as i only needed one so i can just mess with a quadratic
Lagrange, yes...
Not the right idea huh?
You need to NotThink a bit.
...ok? I do that a lot.
Follow those steps that myin showed, youll end with the right answers
I'm just looking at this and my notes, and I have not a clue.
Is u=tanx?
it's more like Let u = tan x then form a polynomial in u, factor it and then substitute back the u =tan x and then solve it ..
Imma do this slowly... If yo ugotta do something while you help me, do it now as I work this thru...
IS factor Thereom involved?
yes...if u-2 is a factor of P(u) ...P(2)=0
Let tan (x) = u \[u ^{3}-9u ^{2}+17u-6=0\] \[\left( u-2 \right)\left( u^2 - 7u -3\right) =0\]
wait wait waitt.... And that's also not how I was taught (or shown in the notes). I'll try it out after I'm done my thing tho/.
Sorry about that. My internet went wonky.
And now I divide the thing by u-2 right?
u^2-7u+3 ____________________ u-2| u^3-9u^2+17u-6 -(u^3-2u^2) ------------------- -7u^2+17u-6 -(-7u^2+14u) --------------- 3u-6 -(3u-6) -------- 0
pax did i make a mistake?
our factorizations are a lttle different i have \[\left( u-2 \right)\left( u^2 - 7u +3\right) =0\]
myininaya, I wish If I can give you more medals for your patience.
are you talking about me and my division above? lol
I think just waiting for me to finish up my stuff here. Thanks by the way.
oh i had a thing i was doing lol
sorry u r right. it's +3
ITs +3?
just making sure when it gets late i tend to screw up on little things
like sometimes i don't know how to add or multiply sometimes
we were talking about this: \[\left( u-2 \right)\left( u^2 - 7u +3\right) =0 \]
ok. that scared me for a sec.
yep myininaya I was talking about that :D
so we know we have u=2 we also have \[u=\frac{7 \pm \sqrt{49-4(3)}}{2}\] don't forget to replace u with tan(x) and solve for x
Finally, I'm done the division.
you have \[u-2=0 \text{ or } u^2-7u+3=0\]
we need to use the quadratic formula to solve the second equation
I hope I remember this for the test.
so this is trig class?
i could be wrong but this is too long to put on a test
there might be a similar problem he/she puts there
but shorter
Whoops. I didn't type that...Double post or something.
i could be wrong
sometimes teachers get problems from special books
I'm sleepy guys I'm sure if you need more help NotTim (who we decided was not myininaya earlier ;)lol) , fool and pax will continue to help unless they are are going.
K. Thanks man.
Do I look like a man to you?
Please say no
Are you?
I actually have bad vision and human identification problems.
When I'm walking around I will misidentify everyone as the same person.
Ok. Its cool. No worries.
any1 know what I should do at (m-2)(m^2-7m+3)?
For m^2-7m+3, do I have to do the quadratic formula?
I guess I already know what to do with m-2, so I'm fine there.
ERm, are you people just gunna lurk there? It's fine, just....
find all places that u can be 0, then, recalling that u = tanx, solve for x.
you already found one solution is u=2, you can find the others by using the quadratic formula on the quadratic part of your factoring.
Hey, if one of my answers is 6.54, and I wanna get the other angle with it, what do i Do?
Um, sorry if I sound abrupt, but could anyone please answer my most recent question?

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