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    Hey guys, does anyone know how to solve tan2x = -1? like how do you work out the values of x using this? Thanks!
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    (Original post by Maha1999)
    Hey guys, does anyone know how to solve tan2x = -1? like how do you work out the values of x using this? Thanks!
    That's C2 stuff, you have to tan^-1(-1) and double the range your given to get a set of solutions
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    (Original post by bruh2132)
    That's C2 stuff, you have to tan^-1(-1) and double the range your given to get a set of solutions
    Ahh okay, thanks so much!!
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    (Original post by bruh2132)
    That's C2 stuff, you have to tan^-1(-1) and double the range your given to get a set of solutions
    Wait but when i do tan^-1(-1), i get 2x= -1/4 pi, however the answer in the book I am using says 2x=3/4 pi ?
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    you need to use the double angle formula for tan2x
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    (Original post by heisenbatch)
    you need to use the double angle formula for tan2x
    Ohhh thank you soo muchh!
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    (Original post by heisenbatch)
    you need to use the double angle formula for tan2x
    What? That's not needed!

    (Original post by Maha1999)
    Wait but when i do tan^-1(-1), i get 2x= -1/4 pi, however the answer in the book I am using says 2x=3/4 pi ?
    When you arctan (inverse tan) some value you will get \tan(x) = a \Rightarrow x=\arctan(a) as the principle solution (meaning the closest one to 0). Depending on your interval, there are many more solutions since tan has a period of \pi so you'll get grab more solutions as you keep adding or subtracting \pi from your principle solution. So generally you get x=\arctan(a) + n \pi where n is some integer of your choice (principle solution is given when n=0)

    In your case, notice how -\frac{1}{4}\pi is your principle solution but -\frac{1}{4}\pi + \pi = \frac{3}{4} \pi, with n=1 here, hence the other solution.
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    (Original post by RDKGames)
    What? That's not needed!



    When you arctan (inverse tan) some value you will get \tan(x) = a \Rightarrow x=\arctan(a) as the principle solution (meaning the closest one to 0). Depending on your interval, there are many more solutions since tan has a period of \pi so you'll get grab more solutions as you keep adding or subtracting \pi from your principle solution. So generally you get x=\arctan(a) + n \pi where n is some integer of your choice (principle solution is given when n=0)

    In your case, notice how -\frac{1}{4}\pi is your principle solution but -\frac{1}{4}\pi + \pi = \frac{3}{4} \pi, with n=1 here, hence the other solution.
    Omg thank you so much! I was soo confused ahahah xD Anyways thank you so much for your help, much appreciated!!!
 
 
 
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