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Math Help - Where did i go wrong?

  1. #1
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    Unhappy Where did i go wrong?

    I have a problem that look like this: cos(3x)=1/ sqrt2

    I already got the correct anser but my teacher says i have to give a more detailed description of how i solved it and he says that part of calculatón is wrong, this is my calculation:

    The unit circle value for the equation above is 40 degrees or pi/4 radian. after finding this i make 3x=a to make it easier. So i figured that because sin(a)=cos(a) i can from this prove that a=pi/4+2*n*pi ...ooor a=-pi/4+2*n*pi

    Then we switch back to 3x and in order to get x alone we need to divide all sides by 3 in both solutions, and that leaves us with the final anwser:

    x=pi/12+2*n*pi/3

    x=-pi/12+2*n*pi/3

    What would you recommend i add or do you see anything incorrect??
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  2. #2
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    Re: Where did i go wrong?

    Quote Originally Posted by MathChick View Post
    I have a problem that look like this: cos(3x)=1/ sqrt2

    I already got the correct anser but my teacher says i have to give a more detailed description of how i solved it and he says that part of calculatón is wrong, this is my calculation:

    The unit circle value for the equation above is 40 degrees or pi/4 radian. after finding this i make 3x=a to make it easier. So i figured that because sin(a)=cos(a) i can from this prove that a=pi/4+2*n*pi ...ooor a=-pi/4+2*n*pi

    Then we switch back to 3x and in order to get x alone we need to divide all sides by 3 in both solutions, and that leaves us with the final anwser:

    x=pi/12+2*n*pi/3

    x=-pi/12+2*n*pi/3

    What would you recommend i add or do you see anything incorrect??
    The only thing I see wrong with above is that $\pi/4$ radians is 45, not 40, degrees.

    The answers you give at the bottom are correct.
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  3. #3
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    Re: Where did i go wrong?

    Oh right that was a mistake, but is there anything else that is not enough explained in your opinion?or does it look good ))
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  4. #4
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    Re: Where did i go wrong?

    Quote Originally Posted by MathChick View Post
    Oh right that was a mistake, but is there anything else that is not enough explained in your opinion?or does it look good ))
    well it would be clearer if you just said that

    $3x=\arccos(1/\sqrt{2})=\pm \pi/4 + 2\pi n, ~~n \in \mathbb{Z}$

    $x=\pm \pi/12 + 2\pi n/3, ~~n \in \mathbb{Z}$

    rather than the the "method" you used to figure out $\cos(\pi/4)=1/\sqrt{2}$

    but other than that it looks fine.
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  5. #5
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    Re: Where did i go wrong?

    Quote Originally Posted by MathChick View Post
    The unit circle value for the equation above is 40 degrees or pi/4 radian.
    I am not sure what "the unit circle value" means, but this sentence is most likely wrong, even if you replace 40 with 45.

    Quote Originally Posted by MathChick View Post
    So i figured that because sin(a)=cos(a)
    Why?

    I agree that last part, including the final answer, is correct.
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  6. #6
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    Re: Where did i go wrong?

    Quote Originally Posted by MathChick View Post
    I have a problem that look like this: cos(3x)=1/ sqrt2

    I already got the correct anser but my teacher says i have to give a more detailed description of how i solved it and he says that part of calculatón is wrong, this is my calculation:

    The unit circle value for the equation above is 40 degrees or pi/4 radian. after finding this i make 3x=a to make it easier. So i figured that because sin(a)=cos(a) i can from this prove that a=pi/4+2*n*pi ...ooor a=-pi/4+2*n*pi

    Then we switch back to 3x and in order to get x alone we need to divide all sides by 3 in both solutions, and that leaves us with the final anwser:

    x=pi/12+2*n*pi/3

    x=-pi/12+2*n*pi/3

    What would you recommend i add or do you see anything incorrect??
    I suspect that what bothered your teacher is this. Your conversion from cos to sin is, as has been pointed out above, unnecessary, but it also is invalid.

    Yes $-\ \dfrac{\pi}{2} \le 3x \le \dfrac{\pi}{2}\ and\ cos(3x) = \dfrac{1}{\sqrt{2}} \implies 3x = \pm\ \dfrac{\pi }{4}.$

    But $-\ \dfrac{\pi}{2} \le 3x \le \dfrac{\pi}{2}\ and\ sin(3x) = \dfrac{1}{\sqrt{2}} \implies 3x = \dfrac{\pi }{4}\ and\ 3x \ne -\ \dfrac{\pi }{4}.$
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