Question
solve 2sin^2x+sinx-1=0 in the interval [0, 2pi0
Answers
drwls
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
2y^2 +y -1 = 0
(2y - 1) (y + 1) = 0
y = sin x = -1/2 or -1