Пожалуйста решите задачу даю 20 баллов
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