Решить уравнение sin 3х = cos 2х

2сos(5x/2)cos(x/2)-2sin(5x/2)cos(5x/2)=02cos(5x/2)*(cos(x/2)-sin(5x/2))=0cos(5x/2)=0⇒5x/2=π/2+πn⇒x=π/5+2πn/5,n∈zcos(x/2)-sin(5x/2)=0cos(x/2)-cos(π/2-5x/2)=0-2sin(3x/2-π/4)sin(-x+π/4)=02sin(3x/2-π/4)sin(x-π/4)=0sin(3x/2-π/4)=0⇒3x/2-π/4=πk⇒3x/2=π/4+πk⇒x=π/6+2πk/3,k∈zsin(x-π/4)=0⇒x-π/4=πm⇒x=π/4+πm,m∈z