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Math Help - Trigonometry : Stucked @ (b)

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    Trigonometry : Stucked @ (b)

    Given that cos 3x = 4cos^3x -3cosx,

    (a) Find the exact value of 8cos^310 - 6cos10
    (b) solve the equation 8cos^3x - 6cosx = - 1 for 0< x < 180

    Attempted solution

    (a) 8cos^310 - 6cos10 = 2(cos3(10))
    = 2cos30
    = \sqrt{3}

    (b) since 8cos^3x - 6cosx = 2cos 3x
    2cos3x = -1
    I cosine inversed \frac{-1}{2} but am not clear on what to do after that.
    I know about the positive and negative quadrants.

    However, the 3x is giving me some problems on adding and minus...

    Note that 0< x <180
    Last edited by Punch; January 9th 2010 at 03:58 AM.
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    Quote Originally Posted by Punch View Post
    Given that cos 3x = 4cos^3x -3cosx,

    (a) Find the exact value of 8cos^310 - 6cos10
    (b) solve the equation 8cos^3x - 6cosx = - 1 for 0< x < 180

    Attempted solution

    (a) 8cos^310 - 6cos10 = 2(cos3(10))
    = 2cos30
    = \sqrt{3}

    (b) since 8cos^3x - 6cosx = 2cos 3x
    2cos3x = -1
    I cosine inversed \frac{-1}{2} but am not clear on what to do after that.
    I know about the positive and negative quadrants.

    However, the 3x is giving me some problems on adding and minus...

    Note that 0< x <180
    I take it the angle is measured in degrees...

    8\cos^3{10} - 6\cos{10} = 2(4\cos^3{10} - 3\cos{10})

     = 2\cos{(3\cdot 10)}

     = 2\cos{30}

     = \frac{2\sqrt{3}}{2}

     = \sqrt{3}.


    8\cos^3{x} - 6\cos{x} = -1

    2\cos{3x} = -1

    \cos{3x} = -\frac{1}{2}

    Since cosine is negative in the second and third quadrants...

    3x = \{180 - 60, 180 + 60\} + 360n, n \in \mathbf{Z}

    3x = \{120, 240\} + 360n

    x = \{40, 80\} + 120n

    In the region 0 < x < 180, x = \{40, 80, 160\}.
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    Quote Originally Posted by Prove It View Post
    I take it the angle is measured in degrees...

    8\cos^3{10} - 6\cos{10} = 2(4\cos^3{10} - 3\cos{10})

     = 2\cos{(3\cdot 10)}

     = 2\cos{30}

     = \frac{2\sqrt{3}}{2}

     = \sqrt{3}.


    8\cos^3{x} - 6\cos{x} = -1

    2\cos{3x} = -1

    \cos{3x} = -\frac{1}{2}

    Since cosine is negative in the second and third quadrants...

    3x = \{180 - 60, 180 + 60\} + 360n, n \in \mathbf{Z}

    3x = \{120, 240\} + 360n

    x = \{40, 80\} + 120n

    In the region 0 < x < 180, x = \{40, 80, 160\}.
    Why do u use 180 - 60?
    Isn't cos^-1\frac {-1}{2} = 120?
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    Quote Originally Posted by Punch View Post
    Why do u use 180 - 60?
    Isn't cos^-1\frac {-1}{2} = 120?
    If \cos{x} = \frac{1}{2}, the focus angle is x = 60^{\circ}.

    Since we have -\frac{1}{2}, we know that the angle lies in the second or third quadrants.

    To work out the angle, for the second quadrant it's 180^\circ - x, and for the third quadrant it's 180^\circ + x.
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    hmm and also, how do i know what value to add for the 360n? in <br />
3x = \{120, 240\} + 360n<br />
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    Quote Originally Posted by Prove It View Post
    If \cos{x} = \frac{1}{2}, the focus angle is x = 60^{\circ}.

    Since we have -\frac{1}{2}, we know that the angle lies in the second or third quadrants.

    To work out the angle, for the second quadrant it's 180^\circ - x, and for the third quadrant it's 180^\circ + x.
    ok so this means that i always have to use to acute angle
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    Quote Originally Posted by Punch View Post
    hmm and also, how do i know what value to add for the 360n? in <br />
3x = \{120, 240\} + 360n<br />
    There are 360^\circ in a circle, and n is an integer representing how many times you have gone around the unit circle.
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    Quote Originally Posted by Prove It View Post
    There are 360^\circ in a circle, and n is an integer representing how many times you have gone around the unit circle.
    From the graphs that I have referred to, for example cos3x=1/2 will mean 3 rounds and that I will have to add 360 twice to each answer i will get.

    Thanks you have been a great help although these questions are probably noob to you.
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    Quote Originally Posted by Punch View Post
    From the graphs that I have referred to, for example cos3x=1/2 will mean 3 rounds and that I will have to add 360 twice to each answer i will get.

    Thanks you have been a great help although these questions are probably noob to you.
    No. Read the algebra I posted.

    Since you have 3x, to solve for x you divide EVERYTHING by 3. This includes the 360.

    So you will actually add 120 to each answer you get.
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