Решить уравнение. 1) sin3x=cos2x 2) 2sinx-3cosx=3

1 sin3x-sin(π/2-2x)=0 2sin(x/2-π/4)cos(5x/2+π/4)=0 sin(x/2-π/4)=0⇒x/2-π/4=πn⇒x/2=π/4+πn⇒x=⇒/2+2⇒n,n∈z cos(5x/2+π/4)=π/2+πk⇒5x/2=π/4+πk⇒x=π/10+2πk/5,k∈z 2 2sinx-3cosx-3=0 4sin(x/2)cos(x/2)-3cos²(x/2)+3sin²(x/2)-3sin²(x/2)-3cos²(x/2)=0 4sin(x/2)cos(x/2)-6cos²(x/2)=0 2cos(x/2)*(2sin(x/2)-3cos(x/2))=0 cos(x/2)=0⇒x/2=π/2+πn⇒x=π+2πn,n∈z 2sin(x/2)-3cos(x/2)=0/cos(x/2) 2tg(x/2)-3=0⇒tg(x/2)=1,5⇒x/2=arctg1,5+πk⇒x=2arctg1,5+2πk,k∈z