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Thread: 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

    $\displaystyle cosec_(2x)=1$

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

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

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

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

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

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

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

    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

    $\displaystyle cosec_(2x)=1$

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

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

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

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

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

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

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

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

    The answer is actually

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

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


    What happens when you let $\displaystyle n = 1$?
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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

    $\displaystyle cosec_(2x)=1$

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

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

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

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

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

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

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

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

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

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

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

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


    I went from $\displaystyle \frac{25\pi}{12}$ to $\displaystyle \frac{\pi}{12}$ 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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