Question

2sinx^2 + (2- корень из 2)cos x +корень из двух -2=0
Укажите корни, принадлежащие отрезку [5п/2 ; 7п/2]

Answer 1

а)
2sin²x + (2 - √2)cosx + √2 - 2 = 0

2 - 2cos²x + (2 - √2)cosx + √2 - 2 = 0

2cos²x + (√2 - 2)cosx - √2 = 0

cosx = (2 - √2 ± √(2 - 4√2 + 4 + 8√2))/4 = (2 - √2 ± √(√2 + 2)²)/4 = (2 - √2 ± (√2 + 2))/4 = {1; -√2/2}

cosx = 1 => x = 2πn, n ∈ ℤ

cosx = -√2/2 => x = π ± π/4 + 2πk, k ∈ ℤ

б)
3π ± π/4 = {11π/4; 13π/4}

Ответ: б) 11π/4; 13π/4