Cos^6x+sin^6x=4sin^2 2x (^6- степень) - вопрос №1740259664226 от Abdurauf2004 02.12.2020 15:09

Question

Алгебра: Cos^6x+sin^6x=4sin^2 2x (^6- степень)

Abdurauf2004 · Answer

(cos^2x+sin^2x)(cos^4x+sin^4x-sin^2xcos^2x)=(2sin2x)^2
(sin^2x+cos^2x)-2sin^2xcos^2x-sin^2xcos^2x=(4sinxcosx)^2
1-3sin^2xcos^2x=16sin^2xcos^2x
sin^2xcos^2x=1/19
sinxcosx=+-1/sqrt(19)
sin2x=+-2/sqrt(19)
x=+-1/2arcsin2/sqrt(19)+Пk/2