Thursday
March 23, 2017

Post a New Question

Posted by on Monday, June 9, 2008 at 5:58pm.

solve 2sin^2x+sinx-1=0 in the interval [0, 2pi0

  • trig - , Monday, June 9, 2008 at 6:41pm

    Call sin a new variable y and solve the quadratic equation for y:
    2y^2 +y -1 = 0
    (2y - 1) (y + 1) = 0
    y = sin x = -1/2 or -1

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question