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Math Help - Trig Problem, where am I going wrong?

  1. #1
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    Trig Problem, where am I going wrong?

    Hello everyone, I'm new to these forums. The long story short is I am trying to finish high school via correspondence, and I need to do two math courses, each containing four units. I am on the fourth unit, the first three gave me very few problems. Anyways, here is the problem:

    2cos^2x+cosx=0 Solve for the domain (0, 2pi) to nearest hundredth of a radian.

    I did this:

    Let k=cosx

    2k^2+k
    =k(2k+1) or
    =cosx(2cosx+1)

    cosx=0
    x=cos^-1(0)
    x=1.57 (this is pi/2 or 90degrees)

    CAST RULE:
    pi-1.57=4.71
    pi+1.57=4.71
    2pi-1.57=4.71

    Therefor for the first zero, x=1.57, 4.71

    For the second zero:

    2cosx+1=0
    2cosx=(-1)
    cosx=(-1/2)
    x=cos^-1(-1/2)
    x=2.09 (this is 2pi/3 or 120degrees)

    CAST RULE: since cosx is negative, the angle is in quadrants two and three:

    pi-2.09=1.05
    pi+2.09=5.23

    So we have 1.57, 4.71, 1.05, 5.23
    -------------------------------------

    However when I check my answer on mathway, it shows the solutions as:

    1.57 (correct)
    4.71 (correct)
    2.09 (?)
    4.19 (?)

    I don't understand why my calculations for the second zero are wrong. If I use quadrants ONE and FOUR (where cosx should be positive), I get 2.09 and 4.19, which is what mathway shows as the correct answers.

    Can someone tell me where I'm going wrong here? The second zero calculates to cosx=(-1/2), and cosx is negative in quadrants two and three, not one and four ... I don't see my error.
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  2. #2
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    Hello, Jeev!

    Welcome aboard!


    2\cos^2\!x+\cos x\:=\:0

    Solve for the domain (0,\:2\pi) to nearest hundredth of a radian.

    Factor: . \cos x(2\cos x + 1) \:=\:0

    Then: . \begin{array}{ccccccccccc}\cos x \:=\:0 & \Rightarrow &\boxed{x \:=\:\dfrac{\pi}{2},\:\dfrac{3\pi}{2}} \\ \\[-3mm]<br />
2\cos x + 1 \:=\:0 & \Rightarrow & \cos x \:=\:\text{-}\frac{1}{2} & \Rightarrow &\boxed{ x \:=\:\dfrac{2\pi}{3},\:\dfrac{4\pi}{3}} \end{array}

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  3. #3
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    Thanks for giving me the solutions, but I am not understanding how you get the last two.

    The CAST rule states that when cosx<0, the angle is in quadrants two and three, which would mean:

    (pi) - 2.09
    (pi) + 2.09

    All I need to understand is, for the second zero, if cosx=(-1/2), why are we calculating the angle in quadrants one and four?
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  4. #4
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    Quote Originally Posted by Jeev View Post
    Thanks for giving me the solutions, but I am not understanding how you get the last two.

    The CAST rule states that when cosx<0, the angle is in quadrants two and three, which would mean:

    (pi) - 2.09
    (pi) + 2.09

    All I need to understand is, for the second zero, if cosx=(-1/2), why are we calculating the angle in quadrants one and four?
    When you put the equation into the calculator, it realised that you've said the angle is negative and calculated the angle in the second quadrant for you.

    To do it analytically you need to consider the POSITIVE angle (called the FOCUS angle) and make the adjustments using the CAST (or as I call it, All Students Talk Crap) rule.

    Consider \cos{\theta} = \frac{1}{2}

    Then your focus angle \theta = \arccos{\frac{1}{2}} = \frac{\pi}{3}.

    Now, since we are told the angle is negative, we need to consider the 2nd and 3rd quadrants.

    So our angles are \pi - \frac{\pi}{3} and \pi + \frac{\pi}{3}. This gives us our angles, \frac{2\pi}{3} and \frac{4\pi}{3}.

    Understand now?
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  5. #5
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    Okay, let me see if I got this.

    cosx(2cosx+1)

    For 2cosx+1

    cosx=(-1/2)

    We need to calculate the "focus angle" which is

    x=arccos(1/2)
    x=1.04 (the question wants radians to the nearest hundredth)

    NOW at this point we consider the negative sign, which dictates that the angles are in quadrants II and IV:

    (pi)-1.04=2.10
    (pi)+1.04=4.18

    Is this the right way to go about this?
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  6. #6
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    Quote Originally Posted by Jeev View Post
    Okay, let me see if I got this.

    cosx(2cosx+1)

    For 2cosx+1

    cosx=(-1/2)

    We need to calculate the "focus angle" which is

    x=arccos(1/2)
    x=1.04 (the question wants radians to the nearest hundredth)

    NOW at this point we consider the negative sign, which dictates that the angles are in quadrants II and IV:

    (pi)-1.04=2.10
    (pi)+1.04=4.18

    Is this the right way to go about this?
    Yes, but don't do any rounding until the very last step. Otherwise you get roundoff errors. You said the answers were 2.09 and 4.19 after all...
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  7. #7
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    Then your focus angle .

    How do you get radians as a fraction of pi as opposed to a decimal value? When I type this in my calculator it obviously just gives me the decimal. Is there some sort of conversion factor I can use?
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  8. #8
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    Quote Originally Posted by Jeev View Post
    Then your focus angle .

    How do you get radians as a fraction of pi as opposed to a decimal value? When I type this in my calculator it obviously just gives me the decimal. Is there some sort of conversion factor I can use?
    Haven't you been taught about the special triangles yet?

    Think of an isoceles right angle triangle, with the shortest sides of unit length. Obviously the angles are 45^\circ, 45^\circ, 90^circ.

    Using Pythagoras, the hypotenuse can then be calculated to be \sqrt{2} units.

    Since \sin{45^\circ} = \frac{O}{H} and \cos{45^\circ} = \frac{A}{H}, we find

    \sin{45^\circ} = \cos{45^\circ} = \frac{1}{\sqrt{2}}.

    By rationalising the denominator, we could also write this as \frac{\sqrt{2}}{2}.

    Also, \tan{45^\circ} = \frac{O}{A} = \frac{\sqrt{2}}{\sqrt{2}} = 1.

    Make the appropriate conversions for radian measures.


    Now for the other triangle.

    Think of an equilateral triangle of side length 2 units. Obviously all the angles are 60^\circ.

    Draw a line perpendicular to the base which cuts the triangle in half and bisects the top angle.

    Now we have two 30^\circ, 60^\circ, 90^\circ triangles.

    The hypotenuse of each is 2 units. One of the shorter lengths is 1 unit. By Pythagoras, we find the other shorter length to be \sqrt{3} units.

    What can you tell me about \sin{30^\circ}, \cos{30^\circ}, \tan{30^\circ}, \sin{60^\circ}, \cos{60^\circ} and \tan{60^\circ}? Make appropriate conversions for radian measure.
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  9. #9
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    Ah so the cos60deg = 1/2

    So we are converting 60deg to radian form, so

    1deg=(pi)/180
    60deg=60(pi)/180
    60deg=(pi)/3

    I understand this now, thanks so much!
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