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Thread: 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; Jul 19th 2010 at 12:28 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 $\displaystyle \cos^2 \theta = 1 - \sin^2 \theta$ and re-arrange the resulting into a quadratic equation where $\displaystyle \sin \theta$ 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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    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 $\displaystyle w = \sin \theta$. Then, following from my earlier reply, you have:

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

    Solve for w. One solution is rejected (why?). The other solution leads to $\displaystyle \sin \theta = \frac{1}{2}$. 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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