cosx - sinx = 1 || : √2 || ⇔ (1/√2)*cosx - (1/√2)*sinx = 1/√2 ⇔
cos(π/4)*cosx - sin(π/4)*sinx = 1/√2 ⇔cos(x +π/4) = 1 /√2 ⇔
x +π/4 = ±π/4 +2πk , k ∈ ℤ , т.е. [ x = - π/2 + 2πk , x = 2πk , k ∈ ℤ .
ответ : - π/2 + 2πk ; 2πk , k ∈ ℤ .
Пошаговое объяснение:
cosx - sinx = 1 || : √2 || ⇔ (1/√2)*cosx - (1/√2)*sinx = 1/√2 ⇔
cos(π/4)*cosx - sin(π/4)*sinx = 1/√2 ⇔cos(x +π/4) = 1 /√2 ⇔
x +π/4 = ±π/4 +2πk , k ∈ ℤ , т.е. [ x = - π/2 + 2πk , x = 2πk , k ∈ ℤ .
ответ : - π/2 + 2πk ; 2πk , k ∈ ℤ .
Пошаговое объяснение: