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deanonm
@deanonm
June 2022
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16^cos^2x+16^sin^2x=10 на [-pi/2;pi]
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LyubaAlexandorovna
16^(cos^2x)+16^(1-cos^2x)=10
16^(cos^2x)+16/16^(cos^2x)-10=0
Заменим 16^(cos^2x)=t
t+16/t-10=0 (*t)
t^2-10*t+16=0
t1,2=(10±√(100-64))/2=(10±6)/2
t1=2 16^(cos^2x)=2 2^(4*cos^2x)=2^1 4*cos^2x=1 cos^2x=1/4 cosx=±1/2 x=±П/3+Пn, nЄZ
t2=8 16^(cos^2x)=8 2^(4*cos^2x)=2^3 4*cos^2x=3 cos^2x=3/4
cosx=±√3/2 x=±П/6+Пm, mЄZ
[-П/2; П] х=-П/3; -П/6; П/6; П/3; 2*П/3; 5*П/6
2 votes
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deanonm
Спасибо
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Answers & Comments
16^(cos^2x)+16/16^(cos^2x)-10=0
Заменим 16^(cos^2x)=t
t+16/t-10=0 (*t)
t^2-10*t+16=0
t1,2=(10±√(100-64))/2=(10±6)/2
t1=2 16^(cos^2x)=2 2^(4*cos^2x)=2^1 4*cos^2x=1 cos^2x=1/4 cosx=±1/2 x=±П/3+Пn, nЄZ
t2=8 16^(cos^2x)=8 2^(4*cos^2x)=2^3 4*cos^2x=3 cos^2x=3/4
cosx=±√3/2 x=±П/6+Пm, mЄZ
[-П/2; П] х=-П/3; -П/6; П/6; П/3; 2*П/3; 5*П/6