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prelim question (1 Viewer)

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hey,

this is a question concerning the trig topic in the prelim course.
is there a simpler way to do it rather than polynomials since we havent done that part of the course yet.

3tan^3 x - 2tan^2 x - tanx + 1 = 0 solve between 0 and 360 degrees.

thanks
 

Timothy.Siu

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hey,

this is a question concerning the trig topic in the prelim course.
is there a simpler way to do it rather than polynomials since we havent done that part of the course yet.

3tan^3 x - 2tan^2 x - tanx + 1 = 0 solve between 0 and 360 degrees.

thanks
i dont even know how to do it knowing polynomials lol
 

clintmyster

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have no idea but let tan x = t or something and use all that sum of roots crap till you get something xD or use the cubic formula which is not in our syllabuses.
 
Last edited:
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haha, the Q is acturaly

3tan^3 x - 3tan^2 x - tanx + 1 = 0 solve between 0 and 360 degrees.

the -2tan^2 x changed to -3tan^2 x

sorry , ive got the right answer now anyway.
 

addikaye03

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haha, the Q is acturaly

3tan^3 x - 3tan^2 x - tanx + 1 = 0 solve between 0 and 360 degrees.

the -2tan^2 x changed to -3tan^2 x

sorry , ive got the right answer now anyway.
Let P(x)=3tan^3 x - 3tan^2 x - tanx + 1

So use the common factors of the constant term to find factors ie. 1=1x1 or -1x-1

So when we try P(1) we get 0 .'. (x-1) is a factor of P(x)

_____________________
Use long division: (x-1) ) (3x^3-3x^2-x+1) Where x=tanx

So in the form P(x)=Q(x)A(x)+R(x) =(x-1)(3x^2-1)

NOW, set this factorisation=0 ( from original part of Q)

(x-1)(3x^2-1)=0

x=1 or/ x^2=1/3

x=1 and x=+-1/rt3 ( plus or minus)

I didn't do type out the long division, hopefully you get what i got, that line is supposed to be the long division line thing
 
Last edited:

jchoi

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Those method above are like all polynomials innit? Anyways, I don't think so, (@ original question) except if you can be bothered, try doing double and triple angles, then putting them all over common denominator.

I think though, it would end up giving only a part of the answer.
 

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