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Math Help - New problem: 2cos^2 2x = cos2x + 1

  1. #1
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    Unhappy New problem: 2cos^2 2x = cos2x + 1

    This time with a harder one. All the hints are really appreciated. Been banging my head to wall few days with this one...

    So, here it is:

    2\cos^2 2x = \cos 2x + 1

    Thanks!

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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by thex10 View Post
    This time with a harder one. All the hints are really appreciated. Been banging my head to wall few days with this one...

    So, here it is:

    2\cos^2 2x = \cos 2x + 1

    Thanks!

    --
    When you think you know it, you don't!
    Note that you can rewrite it as 2\cos^22x-\cos 2x-1=0. Also observe that if you make a substitution, say u=\cos 2x, the equation becomes quadratic in form: 2u^2-u-1=0.

    Solve this quadratic equation for u, and then backsubstitute u=\cos 2x to find x.

    Can you take it from here?
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  3. #3
    Member purplec16's Avatar
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    Quote Originally Posted by thex10 View Post
    This time with a harder one. All the hints are really appreciated. Been banging my head to wall few days with this one...

    So, here it is:

    2\cos^2 2x = \cos 2x + 1

    Thanks!

    --
    When you think you know it, you don't!
    What are you suppose to be doin? verifying the identity or findin the solutions?
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  4. #4
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    skeeter's Avatar
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    Quote Originally Posted by purplec16 View Post
    What are you suppose to be doin? verifying the identity or findin the solutions?
    the equation is conditional ... it's not an identity.
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  5. #5
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    Quote Originally Posted by Chris L T521 View Post
    Note that you can rewrite it as 2\cos^22x-\cos 2x-1=0. Also observe that if you make a substitution, say u=\cos 2x, the equation becomes quadratic in form: 2u^2-u-1=0.

    Solve this quadratic equation for u, and then backsubstitute u=\cos 2x to find x.

    Can you take it from here?
    Thank you! Now I understood it

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