Автор - tliosoml245

Решить систему неравенств. помогите пожалуйста cosx*siny=корень2/2 и x+y=3П/4

Ответ

Автор - elena20092

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


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