Sin(3ПX\2 - П\3)= cos(П\6 - ПX)

Question

Sin(3ПX\2 - П\3)= cos(П\6 - ПX)

Answer 1

Sin(3πx/2-π/3)-sin(πx-π/3)=0
2sin(πx/2)cos(5πx/2-2π/3)=0
sin(πx/2)=0⇒πx/2=πn⇒x=2n,n∈Z
cos(5πx/2-2π/3)=0⇒5πx/2-2π/3=π/2+πn⇒5πx/2=7π/6+πn⇒x=7/15+2n/5,n∈Z