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Thread: Solving trigonometry equation

  1. #1
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    Solving trigonometry equation

    Solve the equation cos2 x-sin2x=cosx 0≤x<2π
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by kelsey3 View Post
    Solve the equation cos2 x-sin2x=cosx 0≤x<2π
    Is it $\displaystyle \cos^2x$ or $\displaystyle \cos 2x $ ?

    Is it $\displaystyle \sin^2x$ or $\displaystyle \sin 2x$ ?
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,



    Is it $\displaystyle \cos^2x$ or $\displaystyle \cos 2x $ ?

    Is it $\displaystyle \sin^2x$ or $\displaystyle \sin 2x$ ?
    I was thinking it was $\displaystyle cos2x$ and $\displaystyle sin2x$.... then i got confused with what i was doing....
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  4. #4
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    Its squared
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  5. #5
    Moo
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    Quote Originally Posted by kelsey3 View Post
    Its squared
    For the two ones ?
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  6. #6
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    Quote Originally Posted by kelsey3 View Post
    Its squared
    that makes a difference! thank you! i'll see if i can do anything with that.... i bet Moo can figure it out long before me though!
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  7. #7
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    ya they're both squared
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  8. #8
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    Quote Originally Posted by kelsey3 View Post
    Solve the equation cos2 x-sin2x=cosx 0≤x<2π
    $\displaystyle \cos^2(x)-(1-\cos^2(x))=\cos(x)$

    $\displaystyle 2\cos(x)-\cos(x)-1=0 \iff (2\cos(x)+1)(\cos(x)-1)=0$

    Now we just need to solve

    $\displaystyle 2\cos(x)+1=0$ and

    $\displaystyle \cos(x)-1=0$


    Good luck.
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  9. #9
    Moo
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    Ok.

    $\displaystyle \cos^2x-\sin^2x=\cos(2x)$

    ~~~~~~~~
    Why ?

    We know that $\displaystyle \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

    $\displaystyle \cos(a+a)=\cos(2a)=\cos^2(a)-\sin^2(a)$
    ~~~~~~~~

    The equation to solve is now :

    $\displaystyle \cos(2x)=\cos(x)$


    Now, you should know that if $\displaystyle \cos(a)=\cos(b)$, then $\displaystyle a=b+2k \pi$ or $\displaystyle a=-b+2k \pi$

    Try to find the solutions, it has nothing difficult


    ----------------------------------

    Other method, maybe longer, but I prefer it :

    Remember that $\displaystyle \sin^2 x+\cos^2x=1 \implies \sin^2x=1-\cos^2x$

    The equation is now :
    $\displaystyle \cos^2x-1+\cos^2x=\cos x$

    $\displaystyle 2\cos^2x-\cos x-1=0$

    $\displaystyle (\cos x-1)(2 \cos x+1)=0$

    -> $\displaystyle \cos x-1=0$
    or $\displaystyle 2 \cos x+1=0$



    Edit : too slow
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  10. #10
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    aw moo! you were too slow! I bet it's my fault! hehe....

    Wow, i never would have gotten that.... But I get it now! Thank you Kelsey3 for askin'!
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  11. #11
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    $\displaystyle
    \cos^2{x}-\sin^2{x}=\cos{x}
    $
    $\displaystyle
    \cos^2{x}+\sin^2{x}=1
    $
    So,
    $\displaystyle
    \sin^2{x}=1-\cos^2{x}
    $
    Substitute in the original equation:
    $\displaystyle
    \cos^2{x}-(1-\cos^2{x})=\cos{x}
    $
    $\displaystyle
    \cos^2{x}-1+\cos^2{x}=\cos{x}
    $
    $\displaystyle
    2\cos^2{x}-1=\cos{x}
    $
    $\displaystyle
    2\cos^2{x}-\cos{x}-1=0
    $
    $\displaystyle
    (2\cos{x}+1)(\cos{x}-1)=0
    $
    $\displaystyle
    \cos{x}=-0.5,\;1
    $
    First case:
    $\displaystyle
    \cos{x}=-0.5
    $
    $\displaystyle
    x=\pm\frac{2\pi}{3}+2n\pi
    $
    $\displaystyle
    x=\frac{2\pi}{3},\;\frac{4\pi}{3}
    $
    Second case:
    $\displaystyle
    \cos{x}=1
    $
    $\displaystyle
    x=\frac{\pi}{2}+2n\pi
    $
    $\displaystyle
    x=\frac{\pi}{2}
    $

    Final solution:
    $\displaystyle
    x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac {\pi}{2}
    $

    nadavs
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