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    Hi,

    I'm doing exercise 1C from the core 3 and 4 aqa book and have got stuck on question 8.
    It says f(x)=/tan2x/ where / represents the modulus line
    Solve f(x)=/tan2x/=1 for -90degrees </= x </= +90degrees

    I get the +/- 22.5 degrees but I don't understand how they also get +/- 67.5 degrees.

    Any help would be greatly appreciated.
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    (Original post by Blake Jones)
    Hi,

    I'm doing exercise 1C from the core 3 and 4 aqa book and have got stuck on question 8.
    It says f(x)=/tan2x/ where / represents the modulus line
    Solve f(x)=/tan2x/=1 for -90degrees </= x </= +90degrees

    I get the +/- 22.5 degrees but I don't understand how they also get +/- 67.5 degrees.

    Any help would be greatly appreciated.
    What is your working?

    Remember that if |\tan 2x| = 1 then  \tan 2x = 1 or \tan 2x = -1 .
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    The problem you have is that when you get 2x=+-45 degrees you also have 2x=+-135 degrees (tan function, add/subtract 180) which gives you +-67.5 degrees for x. It's helpful to use a CAST diagram and to rewrite your bounds as -180<=2x<=180
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    Well you would get  2x= 45+360n , \ n\in \mathbb{Z} . That solves  \tan 2x=1 but you also have  2x=-45+360n which solves  \tan 2x =-1 .
    These are general solutions but you just plug values of n in so that the solutions fall in the domain specified. Remember that the tangent function is periodic with a period of 180 degrees.
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    (Original post by Bananapeeler)
    The problem you have is that when you get 2x=+-45 degrees you also have 2x=+-135 degrees (tan function, add/subtract 180) which gives you +-67.5 degrees for x. It's helpful to use a CAST diagram and to rewrite your bounds as -180<=2x<=180
    Ah thank you! I added/subtracted 180 from the 22.5 not the original!
 
 
 
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