Question

пожалуйста решите задачу даю 20 баллов

Answer 1

sin2x-sin(π/2-2x)=√2cos6x
2sin(2x-π/4)cosπ/4=√2cos6x
√2sin(2x-π/4)=√2cos6x
sin(2x-π/4)-sin(π/2-6x)=0
2sin(4x-3π/8)cos(π/8-2x)=0
sin(4x-3π/8)=0
4x-3π/8=πk
4x=3π/8+πk
x=3π/32+πk/4,k∈z
cos(2x-π/8)=0
2x-π/8=π/2+πk
2x=5π/8+πk
x=5π/16+πk/2,k∈z