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Thread: IDK why i'm stuck

  1. #1
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    IDK why i'm stuck

    I'v got this equation: \sin (4x)+\cos (x+45\degree)=0
    so what I did by trig identities is this: \sin (4x)=\sin (x-45\degree)
    And my answers are:
    1. x=-15\degree+120\degree k
    2. x=45\degree+72\degree k

    Now I know my first answer is correct, yet in the second by the book the answer is -27\degree +72\degree k and I don't know how to solve this.
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  2. #2
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    Re: IDK why i'm stuck

    If you take j= k+1, then k= j- 1 and 45+ 72k= 45+ 72(j- 1)= 45- 72+ 72j= -27+ 72j.

    Both answers are correct- they are just indexed differently.
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  3. #3
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    Re: IDK why i'm stuck

    Quote Originally Posted by xbox360 View Post
    I'v got this equation: \sin (4x)+\cos (x+45\degree)=0
    so what I did by trig identities is this: \sin (4x)=\sin (x-45\degree)
    And my answers are:
    1. x=-15\degree+120\degree k
    2. x=45\degree+72\degree k

    Now I know my first answer is correct, yet in the second by the book the answer is -27\degree +72\degree k and I don't know how to solve this.
    $\displaystyle \begin{align*} \sin{ \left( 4\,x \right) } + \cos{ \left( x + \frac{\pi}{4} \right) } &= 0 \\ \sin{ \left( 4\,x \right) } + \cos{ \left( x \right) } \cos{ \left( \frac{\pi}{4} \right) } - \sin{ \left( x \right) } \sin{ \left( \frac{\pi}{4} \right) } &= 0 \\ \sin{ \left( 4\,x \right) } + \frac{1}{\sqrt{2}}\cos{ \left( x \right) } - \frac{1}{\sqrt{2}}\sin{ \left( x \right) } &= 0 \\ 2\sin{ \left( 2\,x \right) } \cos{ \left( 2\,x \right) } + \frac{1}{\sqrt{2}}\cos{ \left( x \right) } - \frac{1}{\sqrt{2}}\sin{ \left( x \right) } &= 0 \\ 2\,\left[ 2\sin{ \left( x \right) } \cos{ \left( x \right) } \right] \left[ \cos^2{ \left( x \right) } - \sin^2{ \left( x \right) } \right] + \frac{1}{\sqrt{2}}\cos{ \left( x \right) } - \frac{1}{\sqrt{2}}\sin{ \left( x \right) } &= 0 \\ 4\sin{ \left( x \right) } \cos{ \left( x \right) } \left[ \cos{ \left( x \right) } - \sin{ \left( x \right) } \right] \left[ \cos{ \left( x \right) } + \sin{ \left( x \right) } \right] + \frac{1}{\sqrt{2}} \left[ \cos{ \left( x \right) } - \sin{ \left( x \right) } \right] &= 0 \\ \left[ \cos{ \left( x \right) } - \sin{ \left( x \right) } \right] \left\{ 4\sin{ \left( x \right) } \cos{ \left( x \right) } \left[ \cos{ \left( x \right) } + \sin{ \left( x \right) } \right] + \frac{1}{\sqrt{2}} \right\} &= 0 \end{align*}$

    Case 1:

    $\displaystyle \begin{align*} \cos{ \left( x \right) } - \sin{ \left( x \right) } &= 0 \\ \cos{ \left( x \right) } &= \sin{ \left( x \right) } \\ 1 &= \frac{\sin{ \left( x \right) } }{\cos{ \left( x \right) }} \\ 1 &= \tan{ \left( x \right) } \\ x &= \frac{\pi}{4} + \pi\,n \textrm{ where } n \in \mathbf{Z} \end{align*}$

    Case 2:

    $\displaystyle \begin{align*} 4\sin{ \left( x \right) } \cos{ \left( x \right) } \left[ \cos{ \left( x \right) } + \sin{ \left( x \right) } \right] + \frac{1}{\sqrt{2}} &= 0 \\ 4\sin{ \left( x \right) } \cos^2{ \left( x \right) } + 4 \sin^2{ \left( x \right) } \cos{ \left( x \right) } + \frac{1}{\sqrt{2}} &= 0 \\ 4\sin{ \left( x \right) } \left[ 1 - \sin^2{ \left( x \right) } \right] + 4\left[ 1 - \cos^2{ \left( x \right) } \right] \cos{ \left( x \right) } + \frac{1}{\sqrt{2}} &= 0 \\ 4\sin{ \left( x \right) } - 4\sin^3{ \left( x \right) } + 4\cos{ \left( x \right) } - 4\cos^3{ \left( x \right) } + \frac{1}{\sqrt{2}} &= 0 \end{align*}$

    Wolfram gives five more solution sets for this...
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    Re: IDK why i'm stuck

    Quote Originally Posted by HallsofIvy View Post
    If you take j= k+1, then k= j- 1 and 45+ 72k= 45+ 72(j- 1)= 45- 72+ 72j= -27+ 72j.

    Both answers are correct- they are just indexed differently.
    I understand the process u did, but I don't know what "j" stand for?
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  5. #5
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    Re: IDK why i'm stuck

    "j" stands for an arbitrary integer just as "k" does! With the relation, as I said, that j= k+ 1.
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  6. #6
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    Re: IDK why i'm stuck

    So technically speaking it is the same answer? If so, Why they write it as the first result of case 2?
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