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