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Math Help - Solving for X - Where am I off?

  1. #1
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    Solving for X - Where am I off?

    If I have tan{x}+1=0, then I have tan{x} = -1

    well, tan is equal to 1 in \frac{\pi}{4}. It's equal to -1 in quadrant 2 and 4.

    So x =  \frac{3}{4}{\pi}+2n{\pi} where n is all reals.

    but why isn't it also x =  \frac{7}{4}{\pi}+2n{\pi}, since tan is negative in the 4th quadrant as well?
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  2. #2
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    Re: Solving for X - Where am I off?

    Actually it's \displaystyle \tan{x} = 1, and the tangent function is positive in the first and third quadrants.

    \displaystyle x = \frac{\pi}{4} + \pi n where \displaystyle n \in \mathbf{Z}.
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    Re: Solving for X - Where am I off?

    Sorry it my quickness to get the latex correct I wrote the problem out wrong. its supposed to have been TanX+1, not tanx-1
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    Re: Solving for X - Where am I off?

    Well to answer your question, both answers are correct. They can be written in a more compact way as \displaystyle x = \frac{3\pi}{4} + \pi n where \displaystyle n \in \mathbf{Z}. The period of the tangent function is \displaystyle \pi, not \displaystyle 2\pi .
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  5. #5
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    Re: Solving for X - Where am I off?

    Quote Originally Posted by DamenFaltor View Post
    Sorry it my quickness to get the latex correct I wrote the problem out wrong. its supposed to have been TanX+1, not tanx-1
    I suppose you still have the same difficulty as before:

    The tangent function is periodic with a period of π not 2π.
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    Re: Solving for X - Where am I off?

    Ah yes, I forgot about the periodicity of Tan being different.

    ok so to provide:

    {x}= \frac{3}{4}{\pi} + {n}{\pi} ; \frac{7}{4}{\pi} + {n}{\pi}

    Would be a correct answer?
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    Re: Solving for X - Where am I off?

    The second of those is redundant, since \displaystyle \frac{3\pi}{4} + \pi = \frac{7\pi}{4}.
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