Предмет: Математика, автор: alen4ik080496

6sin^2(x)-13cos(x)=12

Ответы

Автор ответа: ewgenijbark

sin²x=1-cos²x
6*(1-cos²x)-13cosx=12
6-6cos²x-13cosx-12=0
-6cos²x-13cosx-6=0
6cos²x+13cosx+6=0
cosx=t
6t²+13t+6=0
D=169-4*6*6=25=5²
t1=-38/12<-1, не подходит, т.к. -1≤cosx≤1
t2=1
cosx=1
x=2πn, n∈Z

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