Math (Calc)

Find all solutions to the following equation on the interval 0<=x<=2PI

8cos^2(X)sin^2(X) + 2cos^2(X) - 3 = 0

There are 8 solutions.

If somebody could show me how to do it and not give me the answers, that would be great.

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asked by Bryan
  1. Substitute sin^2(x) = 1-cos^2(x) and expand the expression in powers of
    cos(x), You get a quadratic equation in cos^2(x). So, if you put cos^2(x) = y, the equation is quadratic in y.

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  2. First, solve the #1 Identity for sin^2(x) and plug it into the equation so that we have only cosines in the problem. The #1 Identity if sin^2(x) + cos^2(x) = 1.

    Distribute.

    Factor as you would if the cosine were simply x. Solve for cos(x).

    Then, refer to the unit circle (which should be in your head) to solve for x. Cosine corresponds to the x-coordinate.

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    posted by Michael
  3. I only get 4 answers though. I'm supposed to get 8

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    posted by Bryan
  4. If you put cos(x) = c, the equation becomes:

    8c^4 - 10 c^2 + 3 = 0

    (4 c^2 - 3)(2c^2 - 1) = 0

    c = ±1/2 sqrt(3)

    c = ±1/2 sqrt(2)

    And you see that:

    x = ±pi/6 + n pi

    x = ±pi/4 + n pi

    There are thus 8 solutions per interval of 2 pi.

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  5. ya sorry, I got it confused. Thanks for all the help.

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    posted by Bryan
  6. if you have something like cos^2 x = 1/16
    I am not picking what you actually have of course
    then cos x = 1/4 OR cos x = -1/4
    now cos 4 = 1/4 at 2 places on the unit circle (first quadrant and fourth)
    and
    cos x = -1/4 at two places on the circle (second quadrant and third)

    that gives you four answers
    The other solution for cos^2 gives you four more

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    posted by Damon

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