Ответ:

x=π/4+nπ/2+2kπ, y=π/4+nπ/2, n,k∈Z

Объяснение:

cosxcosy = sin²y

sinxsiny = cos²y

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cosxcosy+sinxsiny = cos²y+sin²y

cos(x-y)=1

x-y=2kπ

x=y+2kπ

sinxsiny = cos²y

sin(y+2kπ)siny = cos²y

sin²y=cos²y

cos²y-sin²y=0

cos2y=0

2y=π/2+nπ

y=π/4+nπ/2

x=π/4+nπ/2+2kπ