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what are the intercepts of the graph of f(x)=4cos^2 -3 on the interval [0, 2pi]?

  • Trig -

    I am sure you meant:
    f(x)=4cos^2 x -3

    for the x-intercepts:
    4cos^2 x -3 = 0
    cos^2 x = 3/4
    cos x = ±√3/2 ---> so x is in all 4 quadrants.
    I know that cos π/6 = √3/2

    so x = π/6, π - π/6, π + π/6, and 2π-π/6
    x = π/6, 5π/6, 7π/6, and 11π/6

    for the y-intercept, let x = 0
    f(0) = 4cos^2 0 - 3
    = 4(1) - 3 = 1

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