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-   -   Solve cos2x-cos^2x-2sinx+3=0 in the interval [0, 2 pi] (https://www.askmehelpdesk.com/showthread.php?t=437655)

  • Jan 21, 2010, 06:23 PM
    rocknnrebellion
    solve cos2x-cos^2x-2sinx+3=0 in the interval [0, 2 pi]
    solve cos2x-cos^2x-2sinx+3=0 in the interval [0, 2 pi]
  • Jan 22, 2010, 06:35 AM
    ebaines

    Hint: use a couple of the usual trig identities to get everything in terms of sin(x). For example:



    Then you can solve for sin(x).

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