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Math Help - general solution of trig equation

  1. #1
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    general solution of trig equation



    Every things is alright now, but the reference answer is n*pi+[(-1)^n-1]*pi/4, which is different from mine, could anyone tell me how to combine that 2 solutions I found together?


    Again, a little problem, I found that the second general solution is strange because it make sec 5x + csc 2x undefined in some cases of n.
    Last edited by piscoau; September 14th 2011 at 07:18 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: general solution of trig equation

    Honestly, I don't understand your question. Is this what you have done? Can you be more specific about what's your solutions etc ... ?
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  3. #3
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    Re: general solution of trig equation

    sorry for that...I have posted the wrong picture, now corrected

    for the first question, I just want to know how to find the solution which is n*pi+[(-1)^n-1]*pi/4, where n is an integer. Is there something wrong with my process?

    for the second question, I think I have solved the equation correctly, but the solution (2n*pi/3 - pi/2) can't satisfy the equation when n=3..etc
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    Re: general solution of trig equation

    Quote Originally Posted by piscoau View Post


    Every things is alright now, but the reference answer is n*pi+[(-1)^n-1]*pi/4, which is different from mine, could anyone tell me how to combine that 2 solutions I found together?
    Hi piscoau,

    For the first problem your answer is correct and the given answer is an equivalent. For,

    x=n\pi+\left[(-1)^n-1\right]\frac{\pi}{4}\mbox{ where }n\in Z

    When n is even, n=2k~;~k\in Z

    x=2k\pi+\left[(-1)^{2k}-1\right]\frac{\pi}{4}=2k\pi\mbox{ where }k\in Z

    When n is odd, n=2k+1~;~k\in Z

    x=(2k+1)\pi+\left[(-1)^{2k+1}-1\right]\frac{\pi}{4}=(2k+1)\pi-\frac{\pi}{2}=2k\pi+\frac{\pi}{2}\mbox{ where }k\in Z

    Hence, x=2k\pi\mbox{ or }x=2k\pi+\frac{\pi}{2}\mbox{ where }k\in Z

    Quote Originally Posted by piscoau View Post
    for the second question, I think I have solved the equation correctly, but the solution (2n*pi/3 - pi/2) can't satisfy the equation when n=3..etc
    Yes you have solved it correctly and it satisfies the equation for n=3.
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