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Math Help - Another easy trig. problem...

  1. #1
    Junior Member jonnygill's Avatar
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    Another easy trig. problem...

    The equation is...

    3sec^2\Theta-4tan\Theta=2

    I substituted \dfrac{1}{cos^2\Theta} in place of sec^2\Theta and \dfrac{sin\Theta}{cos\Theta} in place of tan\Theta.

    I now have...

    \dfrac{3}{cos^2\Theta}-\dfrac{4sin\Theta}{cos\Theta}=2

    I tried cross multiplying to get the left side as one fraction, but it doesn't seem to get me anywhere. It seems like there is some concept i am missing because i end up getting stuck in the same place with a lot of different problems like this.

    note: i am 25 years old and not in high school or college. I am learning trig. because I want to.

    Thanks!
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    Quote Originally Posted by jonnygill View Post
    The equation is...

    3sec^2\Theta-4tan\Theta=2

    I substituted \dfrac{1}{cos^2\Theta} in place of sec^2\Theta and \dfrac{sin\Theta}{cos\Theta} in place of tan\Theta.

    I now have...

    \dfrac{3}{cos^2\Theta}-\dfrac{4sin\Theta}{cos\Theta}=2

    I tried cross multiplying to get the left side as one fraction, but it doesn't seem to get me anywhere. It seems like there is some concept i am missing because i end up getting stuck in the same place with a lot of different problems like this.

    note: i am 25 years old and not in high school or college. I am learning trig. because I want to.

    Thanks!
    \sec^2{x} = 1 + \tan^2{x} ... sub for \sec^2{x} and get a quadratic equation in terms of \tan{x}
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  3. #3
    Junior Member jonnygill's Avatar
    Joined
    Feb 2011
    Posts
    57
    thank you!

    i used the quadratic equation and got 45 degrees and 18.435 degrees.
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