Question

Найдите корни уравнения 1-cos8x=sin4x

Answer 1

$1-cos8x=sin 4x\\ 1-cos^24x+sin^24x=sin4x\\ 1-1+sin^24x+sin^24x=sin4x\\ 2sin^24x-sin4x=0\\ sin4x(2sin4x-1)=0\\ sin4x=0,4x=\pi n,x=\pi n/4\\ sin4x=1/2,4x=(-1)^k\pi/6+\pi k,x=(-1)^l\pi/24+\pi k/4$