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  • Apr 28, 2009, 12:51 PM
    kayladawn
    Trigonometric Equation
    I am having trouble with the following trigonometry problem:

    cos2x= -1/2

    I tried to turn cos2x into 2cos^2x-1 according to the double angle identity, but then I was unable to factor the problem. Where have I gone wrong, or where do I go from here?
  • Apr 28, 2009, 02:13 PM
    Perito

    I'm not sure why you need the identity. Can't you solve it like this or isn't it allowed?







    I guess you could do it this way:







    And now you have two roots and you still have to take the ArcCos, so you haven't gained much.
  • Apr 29, 2009, 08:16 AM
    Unknown008

    My math tell me that



    x= 60 and 120 degrees... for 0 < x < 360

    BUT if you do through identity, x = 30, 60, 120 and 150 for 0 < x < 360
  • Apr 29, 2009, 08:24 AM
    Perito
    Quote:

    Originally Posted by Unknown008 View Post
    My math tell me that



    x= 60 and 120 degrees... for 0 < x < 360

    and if you do through identity, x = 30, 60, 120 and 150 for 0 < x < 360

    I'm confused. Cos(60) = 0.5, not -0.5. You need to be in the 2nd or 3rd quadrant for it to be -0.5. So 2x = ±120 (or 240), x=±60 on 0 < x < 360. Oh. I guess that does give x=120 also. My bad. 30 degrees and 150 degrees certainly don't fit the original problem, however.
  • Apr 29, 2009, 09:31 AM
    kayladawn

    Perito, when I turned in my homework, I found cos x= 1/2 without the identity as you worked it and then I found my solution set, but I got it wrong.
  • Apr 29, 2009, 10:05 AM
    Perito

    Hmmm. :mad: What answer was said to be the correct one?
  • Apr 29, 2009, 10:13 AM
    Unknown008
    Quote:

    Originally Posted by Perito View Post
    I'm confused. cos(60) = 0.5, not -0.5. You need to be in the 2nd or 3rd quadrant for it to be -0.5. So 2x = ±120 (or 240), x=±60 on 0 < x < 360. Oh. I guess that does give x=120 also. My bad. 30 degrees and 150 degrees certainly don't fit the original problem, however.

    Yes,

    cos(60)=0.5

    but

    cos(2x60)=-0.5

    From cos(2x)=-0.5

    And the plus/minus sign for cos(2x)=±0.5 (for the cos^2(x) = 0.25)

    will give the roots in all four quadrants, so

    2x = 60, 120, 240, 300
    x = 30, 60, 120, 150

    :confused:
  • Apr 29, 2009, 10:17 AM
    Perito
    Quote:

    Originally Posted by Unknown008 View Post
    And the plus/minus sign for cos(2x)=±0.5 (for the cos^2(x) = 0.25)

    will give the roots in all four quadrants, so

    2x = 60, 120, 240, 300
    x = 30, 60, 120, 150

    I'm having a tough time with the ±0.5. The original problem only said -0.5. If you go through the trig identity, you get ±0.5, but +5 doesn't fit the original problem so you have to discard it. Let me know where I'm going wrong. :(
  • Apr 29, 2009, 10:40 AM
    Unknown008

    Yeah, I know that those do not fit in the original question... :o was just posting just in case there was a need for passing through trig identity.

    Ok, edited my first post in the thread...
  • Apr 29, 2009, 11:28 AM
    Unknown008

    Quote:

    Perito agrees: Good. I wonder why they marked her wrong.
    Or was it in rads? Then the answers would be



    or

    1.05 , 2.09

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