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March 25, 2017

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solve the following eq. [0<=x<=360]

2cos(3x-72)=1

  • Maths - ,

    Well I haven't done trig in a long time but u believe...
    You divide the 2 to each side to get
    Cos(3x-72)=1/2
    Then do the inverse to both sides to get rid of the cos and end with 3x-72=60 degrees
    So x= 44 degrees
    Cosine is positive in quad1&4
    So 44degrees and 316degrees.

  • Maths - ,

    44&124degrees

  • Maths - ,

    but 3x mean it has turned 360 degree three times, how about the other four ans?

  • Maths - ,

    I plugged in 3x-72=___ (set them equal to 420,660,780,1020)

    Answer:44,124,164,244,284 degrees

  • Maths - ,

    2cos(3x-72) = 1
    cos(x-7) = 1/2
    we know that cos 60° = 1/2 and cos 300° = 1/2
    ( (inverse)cos .5) = 60 )

    3x-72 = 60 or 3x-72=300
    3x = 132 or 3x = 372
    x = 44° or x = 124° ---> your basic angles

    but the period of cos(3x-72) is 360/3 = 120°
    so as long as we don't go beyond your given domain, we can add 120° to any previous answer to get a new answer, so
    x = 44+120 = 164
    x = 164+120 = 284
    x = 284+120 = 404 ---> too big
    x = 124+120 = 244
    x = 244+120364 -- > too big

    so x = 44°, 124°, 164°, 244°, and 284°
    just like L.Bianchessi found for you in the above post

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