Post a New Question

Math

posted by .

Solve 4sin^2x + 4(squareroot of 2)cosx -6 for all real values of x.

  • Math -

    Did you mean

    4sin^2x + 4√2cosx -6 = 0 ?

    if so then
    4(1-cos^2x) + 4√2cosx - 6 = 0
    4 - cos^2x + 4√2cosx - 6 = 0
    2cos^2x + 2√2cosx + 1 = 0

    let cosx = y
    then y = (-2√2 ± √(8 - 4(2)(1))/4
    y = -√2/2

    then cosx = -√2/2

    we know that cos 45 = √2/2

    and x must be in the II or III quadrant, so
    x = 180-45 = 135 degrees or
    x = 180+45 = 225 degrees

    in radians that would be 3pi/4 or 5pi/4 radians

    general solution
    x = 135 + 360k, 225 + 360k, k an integer
    or
    x = 3pi/4 + 2kpi, 5pi/4 + 2kpi

  • Math -

    Thank you for helping me with this problem.

    Joanie

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question