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Math Help - Sinxtanx - tanx = 0 ; Solve for x?

  1. #1
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    Sinxtanx - tanx = 0 ; Solve for x?

    Hey everyone!

    Sinxtanx - tanx = 0 ; Solve for x? 0 < x < 2pi (and including 0 and 2pi, not sure how to input that)

    This is what I have done so far--
    sinxtanx - tanx = 0

    tanx (sinx - 1) = 0

    Set tanx = 0 and sinx - 1 = 0

    tanx = 0 when x = 0, pi, 2pi

    sinx - 1 = 0
    sinx = 1
    x = pi/2

    so x = 0, pi/2, pi, 2pi

    What I am not sure about, is if pi/2 is a solution or not, as I am thinking that tanx is undefined at pi/2.
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  2. #2
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    Re: Sinxtanx - tanx = 0 ; Solve for x?

    \sin(x)\tan(x)-tan(x)=0 \implies \tan(x)[\sin(x)-1]=0 \implies  \frac{\sin(x)}{\cos(x)}\left[\sin(x)-1\right]=0

    You want to find x within your given boundaries where \sin(x)=0 and \sin(x)=1, AS LONG AS \cos(x)\neq0.

    Edit: To answer your apparent question, you are right, x=\tfrac{\pi}{2} is NOT a solution, because \cos(\tfrac{\pi}{2})=0, which cannot be in a denominator.
    Last edited by abender; January 15th 2013 at 11:58 PM.
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    Re: Sinxtanx - tanx = 0 ; Solve for x?

    So because cos(x) = 0 at pi/2, it is not a solution?
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  4. #4
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    Re: Sinxtanx - tanx = 0 ; Solve for x?

    Quote Originally Posted by Pourdo View Post
    So because cos(x) = 0 at pi/2, it is not a solution?
    Correct. Recall that \tan(x)=\frac{\sin(x)}{\cos(x)}.
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