Question

два уравнения 

первое 2sin^2x=3cosx

второе cos2x=cosx

 

 

 

 

 

  

 

 

 

Answer 1

2-2cos^2x=3cosx

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

cosx=(-3+5)/4

cosx=1/2

x=+-П/3+2Пk

2cos^2x-cosx-1=0

cosx=[1+-sqrt(1+8)]/4

cosx=[1+-3]/4

cosx=1                cosx=-1/2

x=2Пk                 x=2/3П+2Пk

                            x=4/3П+2Пk