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Math Help - curious about this

  1. #1
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    curious about this

    Find x , where 0<x<360 , which satisfy

    \sec x +\tan x=4

    \frac{1+\sin x}{\cos x}=4\Rightarrow 1+\sin x=4\cos x

    putting 4\cos x-\sin x=1 and continuing , i found that x=61.9 using r\cos (x+\alpha)=1 where r=\sqrt{17} and \alpha=\tan^{-1}\frac{1}{4}

    But how come it doesn't work when

    i put \sin x-4\cos x= -1 , doing the same thing .. r\sin (x-\alpha)=-1 and i don get the same result , x=61.9 ?

    is it because -1 is not positive ? Is it one of the conditions ?
    Last edited by thereddevils; August 24th 2009 at 03:08 AM.
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    ...
    \frac{1+\sin x}{\cos x}=4\Rightarrow 1+\sin x=4\cos x

    4\cos x-\sin x=1 and continuing , i found that x=61.9

    But how come it doesn't work when

    i put \sin x-4\cos x= -1
    I do not understand what you are saying;
    if you multiply both sides of this
    4\cos x-\sin x=1.0019...


    by -1, after rearranging, you will get this:

    \sin x-4\cos x= -1.0019...
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  3. #3
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    Quote Originally Posted by aidan View Post
    I do not understand what you are saying;
    if you multiply both sides of this
    4\cos x-\sin x=1.0019...


    by -1, after rearranging, you will get this:

    \sin x-4\cos x= -1.0019...
    sorry bout that Aidan , edited .
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