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polinashirokova
@polinashirokova
August 2022
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a)2sin^2 x-корень из 3 sin 2x=0
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sedinalana
Verified answer
2sin²x-√3sin2x=0
2sin²x-2√3sinxcosx=0
2sinx(sinx-√3cosx)=0
sinx=0⇒x=πn.n∈z
sinx-√3cosx=0/cosx
tgx-√3=0⇒tgx=√3⇒x=π/3+πn,n∈z
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Answers & Comments
Verified answer
2sin²x-√3sin2x=02sin²x-2√3sinxcosx=0
2sinx(sinx-√3cosx)=0
sinx=0⇒x=πn.n∈z
sinx-√3cosx=0/cosx
tgx-√3=0⇒tgx=√3⇒x=π/3+πn,n∈z