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Math Help - Trigonometric Functions

  1. #1
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    Trigonometric Functions

    The final line of my attempt contains the 2 answers listed in the book, but I'm not sure what the basis for excluding all the other values under 180^{\circ} is? I have a feeling my approach is slightly incorrect here. Can anyone help me clear this up?

    Many thanks.

    Q.
    Solve the equation \sin2x=\frac{1}{\sqrt{2}} for 0\leq x\leq\pi.

    Attempt: \sin2x=\frac{1}{\sqrt{2}}\rightarrow2x=45^{\circ}. Sin is positive in the 1st & 2nd quadrants.
    \sin2x=\frac{1}{\sqrt{2}}\rightarrow2x=0^{\circ}+4  5^{\circ}=45^{\circ} or 180^{\circ}-45^{\circ}=135^{\circ}
    Since the angle is 2x add 90^{\circ} & 270^{\circ} to each angle (i.e. 2(45^{\circ}) & 2(135^{\circ}):
    2x=45^{\circ},135^{\circ},315^{\circ} or 2x=135^{\circ},225^{\circ},405^{\circ}
    x=22.5^{\circ},67.5^{\circ},157.5^{\circ} or x=67.5^{\circ},112.5^{\circ},202.5^{\circ}

    Ans.
    (From text book): 22.5^{\circ},67.5^{\circ}
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Trigonometric Functions

    It is odd that the stated interval for x is given in radians, yet the roots are given in degrees. Anyway, we are told:

    0\le x\le 180^{\circ} which means:

    0\le 2x\le 360^{\circ}

    From the given equation, we find:

    2x=45^{\circ},\,135^{\circ} hence:

    x=22.5^{\circ},\,67.5^{\circ}
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  3. #3
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    Re: Trigonometric Functions

    Ok, thank you. My only other question then is why, based on the 2 answer outcome to the problem in the first post, is the answer to the similar question of \cos3x=-\frac{1}{2} for 0\leq x\leq180 shown as a three solution answer x=40,80,160?

    Following, on from above I get:
    \cos3x=\frac{1}{2}\rightarrow3x=60. Cos is negative in the 2nd & 3rd quadrants.
    3x=180-60=120 or 3x=180+60=240
    x=40,80
    How do I account for the 160?
    Last edited by GrigOrig99; November 12th 2012 at 11:54 AM.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Trigonometric Functions

    With:

    0\le x\le 180^{\circ}

    we have:

    0\le 3x\le 540^{\circ}

    So, we find:

    3x=120^{\circ},\,240^{\circ},\,480^{\circ} and so:

    x=40^{\circ},\,80^{\circ},\,160^{\circ}
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  5. #5
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    Re: Trigonometric Functions

    Ok, thank you.
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