Question

найдите корни уравнения 1-cos8x=sin4x

Answer 1

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