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Math Help - Funny double-equation problem

  1. #1
    Member wiseguy's Avatar
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    Funny double-equation problem

    5-7sin θ =2cos^2 θ , θ [-360, 450]

    I'm just a little lost with the multiple use of Theda and ...





    Just thought of something... is θ simply another way of writing θ? The sign has no input on what the problem means...?
    Last edited by mr fantastic; July 19th 2010 at 01:28 PM. Reason: Meregd posts, removed quote tags from question.
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  2. #2
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    Quote Originally Posted by wiseguy View Post
    5-7sin θ =2cos^2 θ , θ [-360, 450]

    I'm just a little lost with the multiple use of Theda and ...





    Just thought of something... is θ simply another way of writing θ? The sign has no input on what the problem means...?
    Solutions are required over the interval [-360 degrees, 450 degrees].

    Substitute \cos^2 \theta = 1 - \sin^2 \theta and re-arrange the resulting into a quadratic equation where \sin \theta is the unknown.

    If you need more help, please show all your work and say where you get stuck.

    Also, please don't put questions in quote tags - it makes it too difficult to quote the question when replying.
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  3. #3
    Member wiseguy's Avatar
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    Here's what I got...

    5=2cos^2θ/7sinθ
    (7/2)5=cos^2θ/sinθ
    17.5=cosθ*cotθ

    Can I carry this out like a normal equation? What I use to eliminate the cosx cotx mess?
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  4. #4
    Member wiseguy's Avatar
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    I did an alternative approach to the problem, however I'm not sure if the 1 on the right side works

    5=2cos^2θ/7sinθ
    (7/2)5=1-sin^2θ/sinθ
    17.5=1-sinθ
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  5. #5
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    Quote Originally Posted by wiseguy View Post
    I did an alternative approach to the problem, however I'm not sure if the 1 on the right side works

    5=2cos^2θ/7sinθ
    (7/2)5=1-sin^2θ/sinθ
    17.5=1-sinθ
    Substitute w = \sin \theta. Then, following from my earlier reply, you have:

    5 - 7 w = 2(1 - w^2) \Rightarrow 2w^2 - 7w + 3 =0.

    Solve for w. One solution is rejected (why?). The other solution leads to \sin \theta = \frac{1}{2}. Solve this equation.
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  6. #6
    Member wiseguy's Avatar
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    Okay, I think I got it:

    x=1/2, x=3

    sinθ=1/2, sinθ=3
    arcsin(1/2)=0.523599, arcsin3=no solution

    so there is only one solution, and it is θ=0.523599 ...?

    Thank you
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  7. #7
    Senior Member eumyang's Avatar
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    Quote Originally Posted by wiseguy View Post
    Okay, I think I got it:

    x=1/2, x=3

    sinθ=1/2, sinθ=3
    arcsin(1/2)=0.523599, arcsin3=no solution

    so there is only one solution, and it is θ=0.523599 ...?

    Thank you
    1) θ needs to be in degrees, as stated in the original problem, so θ = 30.

    2) In the future, you should express radian measures as something times pi if you can. IOW it's better to say θ = π/6 instead of 0.523599.... sinθ = 1/2 -> θ = π/6 or 30 is one of those things you should really memorize.
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  8. #8
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    Quote Originally Posted by wiseguy View Post
    Okay, I think I got it:

    x=1/2, x=3

    sinθ=1/2, sinθ=3
    arcsin(1/2)=0.523599, arcsin3=no solution

    so there is only one solution, and it is θ=0.523599 ...?

    Thank you
    Your answer is supposed to be in degrees since the question uses degree.

    arcsin (0.5) = 30 degree and this is one of the solutions in the range given.

    There are more: 150, 390, -330 , -210
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  9. #9
    Member wiseguy's Avatar
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    Okay, how would I tie the thing where sin is positive in the first and second quadrant to the four solutions of 150, 390, -330 , -210?
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  10. #10
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    Quote Originally Posted by wiseguy View Post
    Okay, how would I tie the thing where sin is positive in the first and second quadrant to the four solutions of 150, 390, -330 , -210?
    The reference angle is 30(1st quadrant) , 150(2nd quadrant). Note also that the period of a sin graph is 360. In other words, it repeats itself every 360.

    so 30+360=390

    How about 150+360=510? Look at the range.

    Now you go clockwise direction, where sin is now positive in the 3rd and 4th quadrant.

    In the 3rd quadrant, -(180+30) and the 4th: -(360-30)

    As an alternative, you can use the general formula for sine.
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  11. #11
    Member wiseguy's Avatar
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    Got it!

    Now I have to remember this stuff... lol
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