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Math Help - Solving a trigonometric equation.

  1. #1
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    Solving a trigonometric equation.

    I was asked to solve the following equation:

    2sin(x) + csc(x) = 3

    I solved it by changing the csc(x) to \frac{1}{sin(x)} and then by multiplying through by sin(x), yielding:

    2sin^2(x) - 3sin(x) + 1 = 0

    I then used the "completing the square" method to solve the equation and arrived at sin(x) = 1. Because I ended up with such a neat answer, I'm wondering if I missed some easier way to solve the problem.

    Was there a more simple way I could have solved this that wouldn't have involved completing the square or using the quadratic equation?

    Thanks!
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  2. #2
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    There isn't a simpler method of solution. However, there is another solution to your quadratic. (The solution you found is one solution.)
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  3. #3
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    Oh, did you mean sin(x) = +/- 1?

    Thanks!
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  4. #4
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    Oh, sorry. It should be sin(x) = 1 and sin(x) = 1/2, correct?
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  5. #5
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    Alas, no. If you plug in \sin(x) = -1, it will not satisfy the equation. You know that \sin(x) = 1 is a solution. What do you need to multiply (\sin(x) - 1) by to get 2\sin^{2}(x)-3\sin(x)+1?

    [EDIT] Ja, you got it.
    Last edited by Ackbeet; November 12th 2010 at 02:21 PM. Reason: OPer got it correct.
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  6. #6
    A Plied Mathematician
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    So, now what? For what range of x are you looking for solutions?
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  7. #7
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    0 < or = X < 2 \pi

    So:

    \frac{\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}

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  8. #8
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    I'd say you're correct.
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