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quadcore
@quadcore
July 2022
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sin2x+[tex] \sqrt{3} [/tex]cosx=2sinx+[tex] \sqrt{3} [/tex]
Подробно если можно , зарание спасибо.
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KayKosades
Sin2x+√(3)cosx-2sinx-
√(3)=0
2sinxcosx-2sinx+√(3)cosx-
√(3)=0
2sinx(cosx-1)+
√(3)(cosx-1)=0
(cosx-1)(2sinx+
√(3))=0
cosx=1
x=2pi*n
sinx=-
√(3)/2
x=((-1)^n)*4pi/3+2pi*k
k
∈Z, n∈Z
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Answers & Comments
2sinxcosx-2sinx+√(3)cosx-√(3)=0
2sinx(cosx-1)+√(3)(cosx-1)=0
(cosx-1)(2sinx+√(3))=0
cosx=1
x=2pi*n
sinx=-√(3)/2
x=((-1)^n)*4pi/3+2pi*k
k∈Z, n∈Z