aramireztjca aramireztjca
  • 24-10-2019
  • Mathematics
contestada

Solve on the interval {0,2pie):

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

Respuesta :

alejandrocampos54
alejandrocampos54 alejandrocampos54
  • 24-10-2019

Answer:

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:

Answer Link
emilyperrine22 emilyperrine22
  • 26-03-2021

Answer:

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

Step-by-step explanation:

Answer Link

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