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Вопрос по математике:
Помогите пожалуйста решить тригонометричечкое уравнение 2 cosx+1=0-
Автор:
gilbert
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Ответ:
Пошаговое объяснение: 2 cosx+1=0 cosx=-1/2
х= ±(П-аrccos1/2) +2Пn
х= ±(П-П/3)+2Пn
х= ±2П/3+2Пn, n∈Z
cosx= -1/2
1) cosx= -1/2
x1= 2π/3+ 2πk k∈z
2) cos(2π-x)= -1/2
2π-x= 2π/3+2πk k∈z
x2=4π/3+2πk k∈z
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