Question

sin(2x-$frac{ pi }{6}$)+cos($frac{13 pi }{6}$-2x)=0

Answer 1

sin(2x-π/6)+cos(π/6-2x)=0
sin(2x-π/6) +sin(π/3+2x)=0
2sinπ/12cos(2x-π/4)=0
cos(2x-π/4)=0
2x-π/4=π/2+πn
2x=3π/4+πn
x=3π/8+πn/2