Question

2cos² x- 3cosx- 2>0
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Answer 1

Ответ:

(-7П/6+2Пk;-5П/6+2Пk)

Объяснение:

cosx=t

-1<=t<=1

2t²-3t-2>0

2t²-3t-2=0

t=(3+-√(9+16))/4=(3+-5)/4

t1=2

t2=-1/2

-1<=t<-1/2

cosx<-1/2

(-3П/2+П/3+2Пk;-П/2-П/3+2Пk)

П/3-3П/2=-7П/6

-П/2-П/3=-5П/6

(-7П/6+2Пk;-5П/6+2Пk)