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Math Help - 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 \cos^2x or \cos 2x ?

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



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

    Is it \sin^2x or \sin 2x ?
    I was thinking it was cos2x and 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
    \cos^2(x)-(1-\cos^2(x))=\cos(x)

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

    Now we just need to solve

    2\cos(x)+1=0 and

    \cos(x)-1=0


    Good luck.
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  9. #9
    Moo
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    Ok.

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

    ~~~~~~~~
    Why ?

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

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

    The equation to solve is now :

    \cos(2x)=\cos(x)


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

    Try to find the solutions, it has nothing difficult


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

    Other method, maybe longer, but I prefer it :

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

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

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

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

    -> \cos x-1=0
    or 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
    Newbie nadavs's Avatar
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    <br />
\cos^2{x}-\sin^2{x}=\cos{x}<br />
    <br />
\cos^2{x}+\sin^2{x}=1<br />
    So,
    <br />
\sin^2{x}=1-\cos^2{x}<br />
    Substitute in the original equation:
    <br />
\cos^2{x}-(1-\cos^2{x})=\cos{x}<br />
    <br />
\cos^2{x}-1+\cos^2{x}=\cos{x}<br />
    <br />
2\cos^2{x}-1=\cos{x}<br />
    <br />
2\cos^2{x}-\cos{x}-1=0<br />
    <br />
(2\cos{x}+1)(\cos{x}-1)=0<br />
    <br />
\cos{x}=-0.5,\;1<br />
    First case:
    <br />
\cos{x}=-0.5<br />
    <br />
x=\pm\frac{2\pi}{3}+2n\pi<br />
    <br />
x=\frac{2\pi}{3},\;\frac{4\pi}{3}<br />
    Second case:
    <br />
\cos{x}=1<br />
    <br />
x=\frac{\pi}{2}+2n\pi<br />
    <br />
x=\frac{\pi}{2}<br />

    Final solution:
    <br />
x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac {\pi}{2}<br />

    nadavs
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