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Math Help - Trig Equation

  1. #1
    Member ~berserk's Avatar
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    Trig Equation

    Given the equation 2\cos^2x-\cos x-1=0 find the exact values of x on the interval [0,2\pi)

    well i tried to take out a \cos x to come up with \cos(\cos^2x-1) and i know that the Pythagorean identity is \sin^2x=1-\cos^2x could anyone help steer me in the right direction
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  2. #2
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    Quote Originally Posted by ~berserk View Post
    Given the equation 2\cos^2x-\cos x-1=0 find the exact values of x on the interval [0,2\pi)

    well i tried to take out a \cos x to come up with \cos(\cos^2x-1) and i know that the Pythagorean identity is \sin^2x=1-\cos^2x could anyone help steer me in the right direction
    Hi bers,

    this is a quadratic from one viewpoint.

    cosx=y

    2y^2-y-1=0

    factor

    (2y+1)(y-1)=0
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  3. #3
    Member ~berserk's Avatar
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    I have another one: 3\tan^2(2x)=1 on the same interval. I'm not very good at getting these started because I need to become better at recognizing identities if they are in the problem.
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  4. #4
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    Quote Originally Posted by ~berserk View Post
    I have another one: 3\tan^2(2x)=1 on the same interval. I'm not very good at getting these started because I need to become better at recognizing identities if they are in the problem.
    Sometimes you don't need any identity.
    You've just got to regenerate the angles..

    3tan^2(2x)=1

    tan^2(2x)=\frac{1}{3}

    tan2x=\pm\frac{1}{\sqrt{3}}

    Find the angles 2x up to 4{\pi}

    then find x
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  5. #5
    Member ~berserk's Avatar
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    Well at least it's good to know i was heading in the right direction with it before hand, it's just good to have reassurance. Would i need to use the double angle identity at all: \tan 2x=\frac{2\tan x}{1-\tan^2x} and if so how would i implement it.
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  6. #6
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    Quote Originally Posted by ~berserk View Post
    Well at least it's good to know i was heading in the right direction with it before hand, it's just good to have reassurance. Would i need to use the double angle identity at all: \tan 2x=\frac{2\tan x}{1-\tan^2x} and if so how would i implement it.
    Yes, you could and you'd get another quadratic...

    tan2x=\frac{2tanx}{1-tan^2x}\Rightarrow\ 1-tan^2x=\pm2\sqrt{3}tanx\Rightarrow\ tan^2x\pm2\sqrt{3}tanx-1=0

    and you've got to solve "both" quadratics.
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