Post a New Question

math

posted by .

Solve 3sin2x - 1 = 0

Interval: [0, 2 pi]

Solve 6cos^2x + 5cosx - 6 = 0

  • math -

    For the first one, sin2x = 1/3
    2x = 0.33984, 2.80176 or 6.62302 radians (and one more less than 4 pi)
    x = 0.16992, 1.4088, or 3.3115
    (and one more less than 2 pi that I will leave you to figure out)

    For the second question, factor into
    (2cosx +3)(3cosx -2) = 0
    Then set each factor = 0 and solve for cos x.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question