Question

sin3x-sin7x=√3 sin2x

Answer 1

$sin3x-sin7x=sqrt{3}sin2x\2sinfrac{3x-7x}{2}*cosfrac{3x+7x}{2}=sqrt{3}sin2x\-2sin2x*cos5x-sqrt{3}sin2x=0\-sin2x(2cos5x+sqrt{3})=0\-sin2x=0\2x=pi*n, nin Z\x=frac{pi*n}{2}, nin Z\ili\2cos5x+sqrt{3}=0\cos5x=-frac{sqrt{3}}{2}\5x=бarccos(-frac{sqrt{3}}{2})+2pi*k, kin Z\5x=pm(pi - frac{pi}{6})+2pi*k, kin Z\x=бfrac{pi}{6}+frac{2pi*k}{5}, kin Z$