Решите Уравнение , срочно! sin(2x+pi/6)=cos3x

sin(2x+π/6)-sin(π/2-3x)=02sin((5x/2-π/6)cos(π/3-x/2)=02sin(5x/2-π/6)sin(x/2+π/6)=0sin(5x/2-π/6)=0⇒5x/2-π/6=πn⇒5x/2=π/6+πn⇒x=π/15+2πn/5,n∈zsin(x/2+π/6)=0⇒x/2+π/6=πk⇒x/2=-π/6+πk⇒x=-π/3+2πk,k∈z