Question

Solve on the interval {0,2pie):

(cos x+1)(2cos^2x-3cos x-2)=0

Answer 1

x=pi, x=2pi/3, x=4pi/3

Explanation:

Answer 2

Step-by-step explanatio(cosx+1)(2cos^2x-3cosx-2)=0

set each factor equal to 0

cosx+1=0

cosx=-1

cos is -1 at x=∏=180º

2cos^2x-3cosx-2=0

factor

(2cosx+1)(cosx-2)=0

set each factor equal to 0

2cosx+1=0

2cosx=-1

cosx=-1/2

cos is -1/2 at x=2∏/3=120º and at x=4∏/3=240º

cosx-2=0

cosx=2

the range of cos is [-1,1] which doesn't include 2n: