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Math Help - Finding exact solutions

  1. #1
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    Exclamation Finding exact solutions



    For a) this is my work:
    cos 2θ + cos θ = 0
    2 cos^2θ - 1 + cos θ = 0
    2cos^2θ + 1 cos θ = 1

    1) cos θ = 1
    2) 2cos θ + 1 = 1
    cos θ = 0

    I got those two equations above from factoring out the problem. However The answer in the back of the book says, 60, 180, 300 degrees. I know that what I got for an answer will not come out with those degrees. So can you help find what I did wrong? Thank you so much.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post


    For a) this is my work:
    cos 2θ + cos θ = 0
    2 cos^2θ - 1 + cos θ = 0
    2cos^2θ + 1 cos θ = 1

    1) cos θ = 1
    2) 2cos θ + 1 = 1
    cos θ = 0

    I got those two equations above from factoring out the problem. However The answer in the back of the book says, 60, 180, 300 degrees. I know that what I got for an answer will not come out with those degrees. So can you help find what I did wrong? Thank you so much.
    Only this part of your work is good:
    cos^{2}\theta + cos\theta = 0

    2 cos^{2}\theta - 1 + cos\theta = 0

    2cos^2\theta +cos\theta - 1 = 0............(1)

    you need to find the value of cos\theta. For that, suppose x=cos\theta in equation (1) to get:

    2x^2+ x - 1 =0

    Solve this quadratic equation for x, which is cos\theta
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  3. #3
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    Thank you for another yet excellent and helpful post.

    I had figured I made a mistake in typing up my initial problem solving. It was actually:

    cos 2θ + cos θ = 0
    2 cos^2θ - 1 + cos θ = 0
    2cos^2θ + cos θ = 1 (move the 1 over to the right side)
    cosθ(2cosθ + 1) = 1 (factor out the left side)

    And thus we would have
    1) cos θ = 1
    2) 2cos θ + 1 = 1
    cos θ = 0

    But it would have given a wrong answer none the less.

    I am wondering how come we leave the 1 on the left side and then factor out the left side and set the right side 0? I thought we would have to move over the 1 to the right side first and then factor out the left side and set it equal to 1...
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post
    Thank you for another yet excellent and helpful post.

    I had figured I made a mistake in typing up my initial problem solving. It was actually:

    cos 2θ + cos θ = 0
    2 cos^2θ - 1 + cos θ = 0
    2cos^2θ + cos θ = 1 (move the 1 over to the right side)
    cosθ(2cosθ + 1) = 1 (factor out the left side)

    And thus we would have
    1) cos θ = 1
    2) 2cos θ + 1 = 1
    cos θ = 0

    THIS IS WRONG
    But it would have given a wrong answer none the less.

    I am wondering how come we leave the 1 on the left side and then factor out the left side and set the right side 0? I thought we would have to move over the 1 to the right side first and then factor out the left side and set it equal to 1...
    Without moving the 1 to the right side, you have:

    2cos^{2}\theta +cos\theta - 1 = 0

    Let u = cos\theta , which gives you:

    2x^2+x-1=0

    This is a quadratic equation fo the form ax^2+bx+c=0 where

    a = 2; b=1; c=-1

    x is given by \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}


    Now solve for x. Solving for x will give you the value of

    cos\theta. From there, you can find out what the value(s) of

    \theta is(are).

    First of all try finding x from the quadratic equation, which will be equal to

    cos\theta
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