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    Can someone explain why 2tan2x - tanx = 6 factorises to (2tanx + 3) (tanx-2)

    I understand that 3 and 2 were used because they multiply to make 6, but why doesn't (2tanx)(tanx) = 2tan2x2 instead of 2tan2x ?
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    (Original post by jessyjellytot14)
    Can someone explain why 2tan2x - tanx = 6 factorises to (2tanx + 3) (tanx-2)

    I understand that 3 and 2 were used because they multiply to make 6, but why doesn't (2tanx)(tanx) = 2tan2x2 instead of 2tan2x ?
    Simply a mathematical convention. (tan(x))^2 = tan^2 (x).
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    (Original post by jessyjellytot14)
    Can someone explain why 2tan2x - tanx = 6 factorises to (2tanx + 3) (tanx-2)

    I understand that 3 and 2 were used because they multiply to make 6, but why doesn't (2tanx)(tanx) = 2tan2x2 instead of 2tan2x ?
    because  (tan x )^2 = (tan^2 x)

    if you still don't understand this, you can always say let tan x = y
    then  2y^2 - y - 6 = 0
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    (Original post by jessyjellytot14)
    Can someone explain why 2tan2x - tanx = 6 factorises to (2tanx + 3) (tanx-2)

    I understand that 3 and 2 were used because they multiply to make 6, but why doesn't (2tanx)(tanx) = 2tan2x2 instead of 2tan2x ?
    To make it much easier and clearer, use tanx=y and so you'd have 2y^2 - y - 6 = 0

    I love trig so much, I thought I was going to hate it but now I understand it - it's one of my favourite topics!
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    (Original post by jamestg)
    To make it much easier and clearer, use tanx=y and so you'd have 2y^2 - y - 6 = 0

    I love trig so much, I thought I was going to hate it but now I understand it - it's one of my favourite topics!
    Yes, y-substitution might be the better bet here. Simply solve for y, OP, and then solve back for x using x = tan^-1 (y).
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    (Original post by jessyjellytot14)
    Can someone explain why 2tan2x - tanx = 6 factorises to (2tanx + 3) (tanx-2)

    I understand that 3 and 2 were used because they multiply to make 6, but why doesn't (2tanx)(tanx) = 2tan2x2 instead of 2tan2x ?
    I think you've misinterpreted the meaning of tan

    tan is a function. Not a term itself.

    Let's say you have a function of x called x+1. When you square that, that becomes (x+1)2.

    Similarly, tan(x) is a function of x. It means we perform something on the x to get a whole new term. In fact, this means that because it's a single term, you could replace tan(x) with u. That would mean you'd get from (2tanx)(tanx) to (2u)(u) which you know equals 2u2 or 2(tan(x))2. Now as an add on, it's just a convention that we write (tan(x))n as tann(x)
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    Firstly, it must be noted that:
    tan \: x \neq tan \times x so tan^2 \: x \neq tan^2 \times x

    In actual fact:
    tan^2 \: x = (tan \: x)^2

    As Student403 mentioned, it is very important to understand that tan \: x is a function of x. Writing tan alone is completely meaningless as it is not a term in itself but a function which must have an input x in order to have an output y.

    To help you visualise the problem, if you let  u = tan \: x then your first equation would become 2u^2 - u = 6. If you factorise this and put tan\:x back in, you will arrive at the correct answer.

    2u^2 - u - 6 = 0

    (2u + 3)(u - 2) = 0

    (2tan \:x + 3)(tan \:x - 2) = 0

    Hope that helps
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    (Original post by jamestg)
    To make it much easier and clearer, use tanx=y and so you'd have 2y^2 - y - 6 = 0

    I love trig so much, I thought I was going to hate it but now I understand it - it's one of my favourite topics!
    (Original post by LelouchViRuge)
    because  (tan x )^2 = (tan^2 x)

    if you still don't understand this, you can always say let tan x = y
    then  2y^2 - y - 6 = 0
    Oh okay thank you, that makes so much more sense now!

    Say if you had 8sin22x - 2sin2x - 3 = 0 , and you let sinx= y, how would you express y?

    Because 8sin2x would then be 8y2 but what about 8sin22x ?
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    (Original post by jessyjellytot14)
    Oh okay thank you, that makes so much more sense now!

    Say if you had 8sin22x - 2sin2x - 3 = 0 , and you let sinx= y, how would you express y?
    4y^2 - y - 3 = 0 I think??? Zacken

    I have only had to factorise when the only extra numbers are coefficients
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    (Original post by jessyjellytot14)
    Oh okay thank you, that makes so much more sense now!

    Say if you had 8sin22x - 2sin2x - 3 = 0 , and you let sinx= y, how would you express y?

    Because 8sin2x would then be 8y2 but what about 8sin22x ?
    In this case, you'd let y = \sin 2x to get 8y^2 - 2y - 3=0, where y = \sin 2x \neq \sin x.

    This would then get you a quadratic in y that you can solve and back-substitute y = \sin \mathbf{2}x.

    In general, if you are given \alpha \sin^2 ax + \beta \sin ax + \gamma then make the substitution y= \sin ax.
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    (Original post by Zacken)
    In this case, you'd let y = \sin 2x to get 8y^2 - 2y - 3=0, where y = \sin 2x \neq \sin x.

    This would then get you a quadratic in y that you can solve and back-substitute y = \sin \mathbf{2}x.

    In general, if you are given \alpha \sin^2 ax + \beta \sin ax + \gamma then make the substitution y= \sin ax.
    Ahhhhhh it makes so much sense, thank you!
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    (Original post by jamestg)
    Ahhhhhh it makes so much sense, thank you!
 
 
 
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