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Math Help - [SOLVED] Help with Trigonometric Equations

  1. #1
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    [SOLVED] Help with Trigonometric Equations

    Hi, no matter what I try, I cannot solve these. I keep on ending up with 4 terms which I cannot factor. Well anyway here are the 2 questions. Please show me how to do them.

    Solve for x on the interval 0 degree </= x </= 360 degrees.
    Question A) 2sinxtanx - tanx - 2sinx + 1 = 0
    Question B) cosxtanx - 1 + tanx - cosx = 0

    I just need to get to part where I factor.

    Thanks!
    Last edited by demo1; May 10th 2009 at 07:41 AM.
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  2. #2
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    Quote Originally Posted by demo1 View Post
    Hi, no matter what I try, I cannot solve these. I keep on ending up with 4 terms which I cannot factor. Well anyway here are the 2 questions. Please show me how to do them.

    Solve for x on the interval 0 degree </= x </= 360 degrees.
    Question A) 2sinx - tanx - 2sinx + 1 = 0
    Question B) cosxtanx - 1 + tanx - cosx = 0

    I just need to get to part where I factor.

    Thanks!
    Hi

    Are you sure about A ? 2sinx cancel ... It remains -tanx + 1 = 0

    For B : \cos x \tan x - 1 + \tan x - \cos x = 0 \iff \sin x - 1 + \frac{\sin x}{\cos x} - \cos x = 0 \iff \sin x - \cos x + \frac{\sin x - \cos x}{\cos x} = 0 \iff (\sin x - \cos x) \left(1 + \frac{1}{\cos x}\right) = 0

    I think that you can finish
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  3. #3
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    Thanks, I made a mistake copying the question. I fixed it now!
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  4. #4
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    Quote Originally Posted by running-gag View Post
    Hi

    Are you sure about A ? 2sinx cancel ... It remains -tanx + 1 = 0

    For B : \cos x \tan x - 1 + \tan x - \cos x = 0 \iff \sin x - 1 + \frac{\sin x}{\cos x} - \cos x = 0 \iff \sin x - \cos x + \frac{\sin x - \cos x}{\cos x} = 0 \iff (\sin x - \cos x) \left(1 + \frac{1}{\cos x}\right) = 0

    I think that you can finish
    I think i know how to solve (1/cosx - 1) for the angle but how do I solve (sinx - cosx)?
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  5. #5
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    Quote Originally Posted by demo1 View Post
    Thanks, I made a mistake copying the question. I fixed it now!
    OK

    2 \sin x \tan x - \tan x - 2 \sin x + 1 = 0 \iff (2\sin x - 1) \tan x - (2\sin x - 1) = 0  \iff (2\sin x - 1) (\tan x - 1) = 0
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  6. #6
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    Quote Originally Posted by demo1 View Post
    I think i know how to solve (1/cosx - 1) for the angle but how do I solve (sinx - cosx)?
    \sin x - \cos x = \sqrt{2} \:\left(\frac{\sqrt{2}}{2}\:\sin x - \frac{\sqrt{2}}{2}\:\cos x\right)

    \sin x - \cos x = \sqrt{2} \:\left(\cos\left(\frac{\pi}{4}\right)\:\sin x - \sin\left(\frac{\pi}{4}\right)\:\cos x\right)

    \sin x - \cos x = \sqrt{2} \:\sin\left(x-\frac{\pi}{4}\right)
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  7. #7
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    Wow thanks running-gag! I think I get it now.
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