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Math Help - help: solving equation (cos^2x)(sin^2x)=0

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    help: solving equation (cos^2x)(sin^2x)=0

    i need some help resolvong \cos^2 x \sin^2 x = 0 over the interval [0 degrees and 360 degrees]
    Last edited by mr fantastic; April 14th 2009 at 07:48 PM. Reason: Clarified the latex and fixed a typo in the equation
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  2. #2
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    Quote Originally Posted by darkmoon123 View Post
    i need some help resolvong \cos^2 x \sin^2 x = 0 over the interval [0 degrees and 360 degrees]
    So either \sin x = 0 or \cos x = 0.

    Solve these two equations over the given domain.
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    I think i have the answer but i am not sure, can someone check this:


    (cos^2x)(sin^2x)=0



    (1-sin^2x ) = 0  (1-cos^2x) = 0

    sin^2x=1 square root both sides cos^2x=1 square root both sides

    sinx=1 cosx=1

    {90 degrees, 0 degrees} ?
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  4. #4
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    Quote Originally Posted by darkmoon123 View Post
    I think i have the answer but i am not sure, can someone check this:


    (cos^2x)(sin^2x)=0



    (1-sin^2x ) = 0  (1-cos^2x) = 0

    sin^2x=1 square root both sides cos^2x=1 square root both sides

    sinx=1 cosx=1

    {90 degrees, 0 degrees} ?
    Did you read what I posted?

    And you have only found solutions over [0 degrees, 90 degrees]. And what about the other solutions?
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  5. #5
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    Quote Originally Posted by darkmoon123 View Post
    (cos^2x)(sin^2x)=0

    (1-sin^2x ) = 0  (1-cos^2x) = 0

    sin^2x=1 square root both sides cos^2x=1 square root both sides

    sinx=1 cosx=1
    If you want to use this method, instead of the one explained earlier, you'll need to use what you learned back in algebra about solving quadratic equations.

    In particular, if 1\, -\, \sin^2(x)\, =\, 0, then (1\, -\, \sin(x))(1\, +\, \sin(x))\, =\, 0, so \sin(x)\, =\, \pm 1. And so forth.

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