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Math Help - [SOLVED] Quadratic Trigonometric Equations

  1. #1
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    [SOLVED] Quadratic Trigonometric Equations

    Hi there, I'm new. I'd really like to get a tutor's help with this, but I can't wait until Monday, so I hope you won't mind my asking.

    Anyway, this has got me really stumped. The book is no help; all the examples are much less complicated. I'll just post two examples I'm having trouble with for now.

    In this section I'm supposed to find all values for \theta:

    1.) cos (2\theta) + 6 sin^2\theta = 4

    What I tried to do was to move the 4 to the left side, convert to all sine values, and plug the values into the quadratic formula. I ended up with \frac {\sqrt {15}}{5} and -\frac {\sqrt {15}}{5} . This is probably wrong, even though I'm fairly sure I didn't make any algebraic errors (I probably just chose the wrong process).


    2.) tan\theta = 2 sin\theta

    I tried converting to all sin and cos values. I think my mistake was to try to divide both sides by sin\theta after that, since dividing by a variable is a no-no. I ended up with cos\theta = \frac {1}{2} , which resulted in two correct answers (as special angles), but there are supposed to be four.

    Any help is greatly appreciated and will result in my abject adoration, and cake.
    Last edited by Zipo; November 14th 2009 at 10:07 PM. Reason: Cleaning up math text
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  2. #2
    MHF Contributor
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    West Malaysia
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    Quote Originally Posted by Zipo View Post
    Hi there, I'm new. I'd really like to get a tutor's help with this, but I can't wait until Monday, so I hope you won't mind my asking.

    Anyway, this has got me really stumped. The book is no help; all the examples are much less complicated. I'll just post two examples I'm having trouble with for now.

    In this section I'm supposed to find all values for \theta:

    1.) cos (2\theta) + 6 sin^2\theta = 4

    What I tried to do was to move the 4 to the left side, convert to all sine values, and plug the values into the quadratic formula. I ended up with \frac {\sqrt {15}}{5} and -\frac {\sqrt {15}}{5} . This is probably wrong, even though I'm fairly sure I didn't make any algebraic errors (I probably just chose the wrong process).


    2.) tan\theta = 2 sin\theta

    I tried converting to all sin and cos values. I think my mistake was to try to divide both sides by sin\theta after that, since dividing by a variable is a no-no. I ended up with cos\theta = \frac {1}{2} , which resulted in two correct answers (as special angles), but there are supposed to be four.

    Any help is greatly appreciated and will result in my abject adoration, and cake.
    For (1) , remember that \cos 2\theta=1-2\sin^2 \theta

    (2) By dividing both sides by \sin \theta , you are ignoring the fact that \sin \theta=0 , and that's why there are 2 missing solutions .

    \frac{\sin \theta}{\cos \theta}=2\sin \theta

    \sin \theta=2\sin \theta\cos \theta

    \sin \theta(2\cos \theta-1)=0
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  3. #3
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    Thanks, that helped! In 1.) I was trying to use sin^2 \theta + cos^2 \theta = 1 , which didn't help even though it got rid of one of the terms.

    So, in 2.) I'm supposed to pull sin\theta out of the parentheses instead of trying to convert it to a cosine. That makes a lot more sense.

    I finished those problems and was able to do the rest of my college homework. Thanks so much!
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