Post a New Question

Trig

posted by .

Solve for x in the interval [0,2pi) sin^2x+2cosx=2

  • Trig -

    sin^2x+2cosx=2
    1 - cos^2x + 2cosx - 2 = 0
    cos^2x - 2cosx + 1 = 0
    (cosx - 1)^2 = 0
    .
    .
    .
    cosx = 1

    Can you take it from there?

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question