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ithkjr
@ithkjr
August 2022
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Решите уравнение пожалуйста! cos2x*cosx-sin2x*sinx=0
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DedushkaBake
Cos2x=1-2sin²x, sin2x=2sinxcosx
(1-2sin²x)cosx-2sinxcosxsinx=0
cosx(1-2sin²x-2sin²x)=0
1) cosx=0, x=π/2+kπ, k∈Z
2)1-4sin²x=0, sin²x=1/4, sinx=+-1/√4=+-1/2
2.1) x=1/2
x=((-1)^n)arsin(1/2)+πn, n∈Z
x=((-1)^n)π/6 +πn
2.2) x=-1/2
x=((-1^n))arsin(-1/2)+πn
x=((-1)^(n+1))π/6+πn
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Answers & Comments
(1-2sin²x)cosx-2sinxcosxsinx=0
cosx(1-2sin²x-2sin²x)=0
1) cosx=0, x=π/2+kπ, k∈Z
2)1-4sin²x=0, sin²x=1/4, sinx=+-1/√4=+-1/2
2.1) x=1/2
x=((-1)^n)arsin(1/2)+πn, n∈Z
x=((-1)^n)π/6 +πn
2.2) x=-1/2
x=((-1^n))arsin(-1/2)+πn
x=((-1)^(n+1))π/6+πn