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VishnevskayaSophie
@VishnevskayaSophie
August 2022
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Решите уравнение:
3 cos^2 х-8 cos х+3=0
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СовушкинДомик
3cos²x-8cosx+3=0
Пусть cosx=t, тогда
3t²-8t+3=0
D=(-8)²-4*3*3=64-36=28
t1,2= (8±√28)/6
t1=(8+2√7)/6= 2(4+√7)/6= (4+√7)/3
t2=(8-2√7)/6= 2(4-√7)/6= (4-√7)/3
cosx1= (4+√7)/3 — не подходит, т.к |cosx|<1
cosx2= (4-√7)/3
x= ±arccos((4-√7)/3) + 2пк
Ответ: x= ±arccos((4-√7)/3) + 2пк
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Answers & Comments
Пусть cosx=t, тогда
3t²-8t+3=0
D=(-8)²-4*3*3=64-36=28
t1,2= (8±√28)/6
t1=(8+2√7)/6= 2(4+√7)/6= (4+√7)/3
t2=(8-2√7)/6= 2(4-√7)/6= (4-√7)/3
cosx1= (4+√7)/3 — не подходит, т.к |cosx|<1
cosx2= (4-√7)/3
x= ±arccos((4-√7)/3) + 2пк
Ответ: x= ±arccos((4-√7)/3) + 2пк