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March 26, 2017

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Finding all Solutions for
2cos(x)-ã3=0

  • Trig - ,

    Is this cosx=1/2sqrt3 ?

  • Trig - ,

    no it's 2cosx-sqrt3=0

  • Trig - ,

    I will read it as
    2cosx - √3 = 0
    cosx = √3/2
    x = 30° , from the standard 30-60-90 triangle

    or x = 330°

    "all solutions " would be
    x = 30 + k(360) or x = 330+k(360)° where k is an integer.

  • Trig - ,

    How do you find the solutions?

  • Trig - ,

    since the standard cosine curve has a period of 360°, any answer you have will repeat itself in the next curve or 360° either to the right or to the left.
    by multiplying 360 by k, k and integer, I am simply adding/or subtracting multiples of 360 to any answer.

    to get my original 30°, as I said, I recognized the rations of the 30-60-90 right-angled triangle, which are 1:√3:2.
    Secondly , since the cosine is positive in the 1st and 4th quadrant, the 30° will be the 1st quadrant solution and 360-30 or 330° would be the 4th quadrant solution.

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