How do you solve #cos^2x + 2cosx-3 = 0# over the interval 0 to 2pi?
1 Answer
Feb 14, 2016
# x= 0 , x = 2pi #
Explanation:
first step is to factor the left side.
# cos^2x + 2cosx - 3 = (cosx + 3 )(cosx - 1 ) # hence (cosx + 3 )(cosx - 1 ) = 0
now : cosx + 3 = 0 , cosx _ 1 = 0
cosx + 3 = 0 → cosx = - 3 has no solution.
and cosx - 1 = 0 → cosx = 1#rArr x = 0 , 2pi #
