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Автор - HanaYochi456

К графику функции f(x)=x^2- (2sqrt3+1/sqrt3)*x +sqrt3+1/sqrt3 в точке с абсциссой х=sqrt3+1/sqrt3 проведена касательная. Найти угол между частью касательной, лежащей в верхней полуплоскости(у>0) и положительным направлением оси Ох. Помогите!

Ответ

Автор - ПAmиDoPkA
1) (sinx+cosx)^2-1=sin^2x+sinxcosx +cos^2x - 1= 2sinx2)sin2x=sqrt3/22x = [tex](-1)^{n+1}   [tex]frac{pi}{2} +2pi nx=  

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