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

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- NotTim

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

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- NotTim

Well I tried putting x=1, I got an error.

- anonymous

\[x \approx3.14159 n+1.10715\]
\[x \approx3.14159 n+0.429998\]
\[x \approx3.14159 n+1.4191\]

- myininaya

first see if you can solve this
\[u^3-9u^2+17u-6=0\]

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## More answers

- JamesJ

Thanks myin; that's the way.

- myininaya

if you try u=2, you will see you have one solution

- anonymous

thats what i wasn gonna say replace tan with a variable

- myininaya

then using synthetic division you can find the other 2 solutions

- NotTim

I dunno what that is.

- myininaya

synethic division+quadratic formula (or perhaps factoring if we have other rational zeros)

- nubeer

u can also urs long division to get factors

- NotTim

Wait a sec guys, I'm trying to understand this all...

- JamesJ

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
\[ u^3ā9u^2+17^uā6 = (u-2)q(u) \]
for some second order polynomial

- JamesJ

make that 17u, not 17^u

- myininaya

lol

- NotTim

What happened to tan?

- JamesJ

i'll let the other chefs take over

- myininaya

we will get back to the tan business after we solve that polynomial=0 for u

- NotTim

Ok, so I'm doing the whole F(#) thing at this point right?

- myininaya

what?

- myininaya

what does F(#) mean?

- NotTim

Looking for the value of x that makes everything 0.

- anonymous

u = 2
u = 1/2 (7-sqrt(37))
u = 1/2 (7+sqrt(37))

- anonymous

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.

- myininaya

right fool

- NotTim

What if I got an error on my calculator?

- NotTim

Useless its tanx (as an entirety) that I using....

- anonymous

Your trying to plug number right into the equation?

- myininaya

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

- NotTim

Lagrange, yes...

- NotTim

Not the right idea huh?

- anonymous

You need to NotThink a bit.

- NotTim

...ok? I do that a lot.

- anonymous

Follow those steps that myin showed, youll end with the right answers

- NotTim

I'm just looking at this and my notes, and I have not a clue.

- NotTim

Is u=tanx?

- myininaya

yes

- NotTim

Ok.

- anonymous

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 ..

- NotTim

ERkk.

- NotTim

Imma do this slowly...
If yo ugotta do something while you help me, do it now as I work this thru...

- NotTim

IS factor Thereom involved?

- PaxPolaris

yes...if u-2 is a factor of P(u) ...P(2)=0

- PaxPolaris

Let tan (x) = u \[u ^{3}-9u ^{2}+17u-6=0\] \[\left( u-2 \right)\left( u^2 - 7u -3\right) =0\]

- NotTim

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/.

- NotTim

Sorry about that. My internet went wonky.

- NotTim

And now I divide the thing by u-2 right?

- myininaya

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

- NotTim

ok

- myininaya

pax did i make a mistake?

- myininaya

our factorizations are a lttle different
i have
\[\left( u-2 \right)\left( u^2 - 7u +3\right) =0\]

- anonymous

myininaya, I wish If I can give you more medals for your patience.

- NotTim

Lol.

- myininaya

are you talking about me and my division above? lol

- NotTim

I think just waiting for me to finish up my stuff here. Thanks by the way.

- myininaya

oh i had a thing i was doing lol

- PaxPolaris

sorry u r right. it's +3

- NotTim

ITs +3?

- myininaya

just making sure
when it gets late i tend to screw up on little things

- myininaya

like sometimes i don't know how to add or multiply sometimes

- myininaya

we were talking about this:
\[\left( u-2 \right)\left( u^2 - 7u +3\right) =0 \]

- NotTim

ok. that scared me for a sec.

- anonymous

yep myininaya I was talking about that :D

- myininaya

lol

- myininaya

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

- NotTim

Finally, I'm done the division.

- myininaya

lol

- myininaya

good

- myininaya

you have
\[u-2=0 \text{ or } u^2-7u+3=0\]

- myininaya

we need to use the quadratic formula to solve the second equation

- NotTim

I hope I remember this for the test.

- myininaya

so this is trig class?

- myininaya

i could be wrong but this is too long to put on a test

- myininaya

there might be a similar problem he/she puts there

- myininaya

but shorter

- NotTim

Whoops. I didn't type that...Double post or something.

- myininaya

i could be wrong

- anonymous

sometimes teachers get problems from special books

- myininaya

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.

- NotTim

K. Thanks man.

- myininaya

Do I look like a man to you?

- myininaya

Please say no

- myininaya

lol

- NotTim

Are you?

- NotTim

I actually have bad vision and human identification problems.

- NotTim

When I'm walking around I will misidentify everyone as the same person.

- myininaya

Ok. Its cool. No worries.

- NotTim

any1 know what I should do at (m-2)(m^2-7m+3)?

- NotTim

For m^2-7m+3, do I have to do the quadratic formula?

- NotTim

I guess I already know what to do with m-2, so I'm fine there.

- NotTim

ERm, are you people just gunna lurk there? It's fine, just....

- anonymous

find all places that u can be 0, then, recalling that u = tanx, solve for x.

- anonymous

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.

- NotTim

Ok.

- NotTim

Hey, if one of my answers is 6.54, and I wanna get the other angle with it, what do i Do?

- NotTim

Hi?

- NotTim

Um, sorry if I sound abrupt, but could anyone please answer my most recent question?

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