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Math Help - Solve Equations 0< or = x < or = 2pi

  1. #1
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    Solve Equations 0< or = x < or = 2pi

    How would you solve these

    Solve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.

    Answear is in form of this
    pi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by j0nath0n3 View Post
    How would you solve these

    Solve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.

    Answear is in form of this
    pi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?
    Note that \sin^2 x=1-\cos^2 x

    So we can rewrite the equation as \cos^2 x -\left[1-\cos^2 x\right]=-\cos x

    This simplifies to 2\cos^2 x-1=-\cos x\implies 2\cos^2x+\cos x-1=0

    Now, if we make the substitution z=\cos x, the equation now becomes the quadratic equation 2z^2+z-1=0\implies (2z-1)(z+1)=0

    This tells us that z=-1 or z=\tfrac{1}{2}

    But z=\cos x, so now that means either \cos x=\tfrac{1}{2} or \cos x=-1

    Now can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Note that \sin^2 x=1-\cos^2 x

    So we can rewrite the equation as \cos^2 x -\left[1-\cos^2 x\right]=-\cos x

    This simplifies to 2\cos^2 x-1=-\cos x\implies 2\cos^2x+\cos x-1=0

    Now, if we make the substitution z=\cos x, the equation now becomes the quadratic equation 2z^2+z-1=0\implies (2z-1)(z+1)=0

    This tells us that z=-1 or z=\tfrac{1}{2}

    But z=\cos x, so now that means either \cos x=\tfrac{1}{2} or \cos x=-1

    Now can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?
    Thats really good and I understood however my answear choices are in radian... and I dont understand why..
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by j0nath0n3 View Post
    Thats really good and I understood however my answear choices are in radian... and I dont understand why..
    In general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is 360\text{ degrees}=2\pi\text{ radians}.
    Last edited by Chris L T521; January 10th 2009 at 10:55 PM. Reason: typo
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  5. #5
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    Re :

    Quote Originally Posted by Chris L T521 View Post
    In general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is 360\text{ degrees}=\pi\text{ radians}.
    Chris , is it a typo , i thought 360\text{ degrees}=2\pi\text{ radians}
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mathaddict View Post
    Chris , is it a typo , i thought 360\text{ degrees}=2\pi\text{ radians}
    Yes, my mistake. Thank you for catching that.
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