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sekimova1 @sekimova1

July 2022 1 2 Report

sinx-√2 sin3x=-sin5x
cos(70°+x)cos(x-20°)=1/2

Answers & Comments

burn0to0die 1) Sinx - √2Sin3x + Sin5x = 02Sin3xCos2x - √2Sin3x = 0Sin3x(Cos2x -√2) = 0Sin3x = 0 или Cos2x -√2= 03x = πn , n∈Z Cos2x =√2x = πn/3, n ∈Z нет решений.2) Сos(70°+x)Cos(x -20°) = 1/2Sin(20° -x)Cos(20° -x) = 1/2 |*22Sin(20° -x)Cos(20° -x) = 1Sin(40° -2x) = 140° -2x = 90° + 360°*n, n ∈Z2x = -50° -360°n, n∈Zx = -25° -180°n, n ∈Z

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