Question

Solve each equation for \[0\le x <2\pi\].
tan ^{2}x-(sqrt3+1)tanx+sqrt3 =0

Answer 1

\[\tan ^{2}x-(\sqrt3+1)tanx+\sqrt3 =0\]

Answer 2

Can you factor this?

Answer 3

I don't know how to factor for this question. :(

Answer 4

to*

Answer 5

see tan(x) --> a for now

Answer 6

What multiplies to make sqrt (3)?

-sqrt(3) and -1?

(tan x - ?)(tan x - ?)

Answer 7

are you sure your expression is right?

Answer 8

I think it is ...

Answer 9

yes.

Answer 10

can you guys show me step by step please?

Answer 11

What about

\[(\tan x - 1)(\tan x - \sqrt{3})\]

Answer 12

that is right. But the question is how did you get there?

Answer 13

a^2-(sqrt(3)+1)a+Sqrt[3]

factored out to be

(a -sqrt[3])(a -1)

Answer 14

a= sqrt[3]

or

a= 1

Answer 15

remember a=tan(x)

Answer 16

I used the trial and error method of factoring from algebra

tan x times tan x = tan^x
-sqrt(3) times -1 = sqrt(3)

Answer 17

tan[x]=sin(x)/cos(x)=sqrt[3]=

sin(60)= sqrt[3]/2
cos(60)= 1/2

sqrt[3]
--------- / ( 1/2) = sqrt[3]
2


hence x=60 degree

Answer 18

I am only having trouble with the factoring in this question.
I still don't get how you guys came up with (tanx-1)(tanx-sqrt3).
I am sorry :(

Answer 19

tan ^{2}x-(sqrt3+1)tanx+sqrt3

for simplicity, I substitute a=tan(x)

a^2-(sqrt[3]+1)a+sqrt[3]

then just factor

Answer 20

Let me help with the factoring ... How would you factor

x^2 - 5x + 6

Answer 21

6 = -3 * -2 which means

( x - 3)(x - 2)??

Answer 22

yep

Answer 23

oddly though i don't know how to factor in this question..

Answer 24

If you know how to factor

a^2 - 5a + 6 = (a - 3)(a-2)

then factor

tan ^{2}x-(sqrt3+1)tanx+sqrt3 =0

can we place the tan x where a is??

Answer 25

yes

Answer 26

And if

-3 and -2 = 6

and

-sqrt(3) and -1 = sqrt(3)

then replace

-3 and -2

with

-sqrt(3) and -1

Answer 27

(x - 3)(x -2)

becomes

(tan x - 1)(tan x - sqrt(3))

Answer 28

success?

Answer 29

you are wonderful! haha but I am still struggling.. mamammia.. :S

Answer 30

one step at a time, you will make it .... the factor formed definitely looks weird and intimidating

Answer 31

thanks! i am keep going over your steps and trying to make the way through! :)