Post a New Question

trig

posted by .

Find all solutions of the equation 2sin^2x-cosx=1 in the interval [0,2pi)

x1= ?
x2=?
x3=?

  • trig -

    2 (1 - cos^2 x) - cos x = 1
    2 - 2 cos^2 x -cos x = 1
    2 cos^2 x + cos x -1 = 0
    let y = cos x
    then
    2 y^2 +y -1 = 0
    (y+1)(2y-1) = 0
    y = .5 or y = -1
    so
    cos x = .5 or cos x = -1
    so
    x = 60 degrees or 300 degrees
    or 180 degrees

  • trig -

    Rewrite as
    2(1-cos^2x)-cosx = 1.
    Rearrange as
    2cos^2x + cos x -1 = 0,
    which can be factored to provide:
    (2cosx -1)(cosx +1) = 0.
    This means
    cosx = 1/2 or -1
    Use that to find the three angles.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question