How do you solve cos^2x + 2cosx-3 = 0cos2x+2cosx−3=0 over the interval 0 to 2pi?

1 Answer
Jim G.
Feb 14, 2016

x= 0 , x = 2pi x=0,x=2π

Explanation:

first step is to factor the left side.

cos^2x + 2cosx - 3 = (cosx + 3 )(cosx - 1 ) cos2x+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 ⇒x=0,2π

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