cosx · siny = √2/2
х + у = 3π/4 → у = 3π/4 -х
сosx · sin(3π/4 - x) = √2/2
сosx · (sin3π/4 · cos x - cos3π/4 ·sinx) = √2/2
cosx · (√2/2 · cosx - (-√2/2) · sinx) = √2/2
cos² + cosx · sinx = 1
cos² + cosx · sinx = sin²x + cos²x
sin²x - sinx · cosx = 0
sinx · (sinx - cosx) = 0
1) sinx = 0 → x1 = πk → y1 = 3π/4 - πk
2) sinx - cosx = 0
cosx ≠ 0 tgx = 1 x2 = π/4 + πk → y2 = 3π/4 - π/4 - πk → y2 = π/2 - πk
Ответ: 1) x1 = πk y1 = 3π/4 - πk
2) x2 = π/4 + πk y2 = π/2 - πk