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Math Help - Solving trig questions

  1. #1
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    Solving trig questions

    Need help on finding out what i have done wrong:
    1)Solve cosec(2x)+1=2
    This is what i have done

    cosec_(2x)=1

    2x=\frac{\pi}{2},\frac{3\pi}{2}

    x=\frac{\pi}{4} and \frac{3\pi}{4}

    Answer says \frac{\pi}{4} and \frac{5\pi}{4}

    2)Solve cot(2x-\frac{\pi}{3})=-1

    2x=\frac{3\pi}{4}+\frac{\pi}{3}, \frac{7\pi}{4}+\frac{\pi}{3}

    2x=\frac{13\pi}{12},\frac{25\pi}{12}

    x=\frac{13\pi}{24},\frac{25\pi}{24}

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Need help on finding out what i have done wrong:
    1)Solve cosec(2x)+1=2
    This is what i have done

    cosec_(2x)=1

    2x=\frac{\pi}{2},\frac{3\pi}{2}

    x=\frac{\pi}{4} and \frac{3\pi}{4}

    Answer says \frac{\pi}{4} and \frac{5\pi}{4}

    2)Solve cot(2x-\frac{\pi}{3})=-1

    2x=\frac{3\pi}{4}+\frac{\pi}{3}, \frac{7\pi}{4}+\frac{\pi}{3}

    2x=\frac{13\pi}{12},\frac{25\pi}{12}

    x=\frac{13\pi}{24},\frac{25\pi}{24}

    P.S
    1) \sin{\frac{3\pi}{2}} = -1, not 1.

    The answer is actually

    2x = \frac{\pi}{2} + 2\pi n

    x = \frac{\pi}{4} + \pi n.


    What happens when you let n = 1?
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    Need help on finding out what i have done wrong:
    1)Solve cosec(2x)+1=2
    This is what i have done

    cosec_(2x)=1

    2x=\frac{\pi}{2},\frac{3\pi}{2}

    x=\frac{\pi}{4} and \frac{3\pi}{4}

    Answer says \frac{\pi}{4} and \frac{5\pi}{4}

    2)Solve cot(2x-\frac{\pi}{3})=-1

    2x=\frac{3\pi}{4}+\frac{\pi}{3}, \frac{7\pi}{4}+\frac{\pi}{3}

    2x=\frac{13\pi}{12},\frac{25\pi}{12}

    x=\frac{13\pi}{24},\frac{25\pi}{24}

    P.S
    \cot{\left(2x - \frac{\pi}{3}\right)} = -1

    \tan{\left(2x - \frac{\pi}{3}\right)} = -1

    2x - \frac{\pi}{3} = \left\{ \frac{3\pi}{4}, \frac{7\pi}{4} \right\} + \pi n

    2x = \left \{ \frac{13\pi}{12}, \frac{25\pi}{12} \right\} + \pi n

     x = \left \{ \frac{13\pi}{24}, \frac{25\pi}{24} \right \} + \frac{\pi n}{2}.
    Last edited by Prove It; January 14th 2010 at 02:52 AM.
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  4. #4
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    <br />
2x = \left \{ \frac{13\pi}{12}, \frac{25\pi}{12} \right\} + \pi n<br />
    why would you want to plus \pi again?
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    why would you want to plus \pi again?
    The period of the tangent function is \pi.
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  6. #6
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    so is this including tan=1 as well, coz is that how you got \frac{\pi}{4}=\frac{\pi}{24}
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  7. #7
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    \frac{\pi}{4}\neq \frac{\pi}{24}.


    I went from \frac{25\pi}{12} to \frac{\pi}{12} because you have gone around the unit circle once.

    Though I may have made a mistake in doing that too early...

    Will edit.
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