Задача 13723 sinx–cos4x=0...

u18813324029

14.02.2017

sinx–cos4x=0

SOVA

14.02.2017

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сos4x=sin((π/2)-4x);

sinx-sin((π/2)-4x)=0;

2sin((x-((π/2)-4x))/2)*cos((x+(π/2)-4x)/2)=0
(5x-(π/2))/2=πk, k∈Z или (3х-( π/2))/2=(π/2)+πn, n∈Z
5x-(π/2)=2πk, k∈Z или 3х-( π/2)=π+2πn, n∈Z

x= (π/10)+(2π/5)k, k∈Z или х=(π/2)+(2π/3)n, n∈Z