Учитель требует объяснений! x - y = π/2 cosx - cosy = - √2 x = π/2 + y cos(π/2 + y) - cosy = - √2 - siny - cosy = - √2 -(cosy + siny) = - √2 - (√2cos(π/4 - y) = - √2 cos(y - π/4) = 1 y - π/4 = 2πk, k∈z y = π/4 + 2πk, k∈z x = π/2 + π/4 + 2πk, k∈z x = 3π/4 + 2πk, k∈z ответ: x = 3π/4 + 2πk, k∈z ; y = π/4 + 2πk, k∈z объясните переход вот к этой строчке: - (√2cos(π/4 - y) = - √2
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Ответы на вопрос:
(cosy+siny) = -cosy - siny = -√2(√2/2*cosy + √2/2 siny) = -√2*((cos(π/4)*cos(y) + sin (π/4)*siny) = -√2*сos(π/4-y) стандартное преобразование √2*√2/2=1
Решим уравнение: cos(2x) = 1 - 2sin^2(x) cos(π/2 - x) = sinx 1 - 2sin^2(x) - sinx -1 = 0 sinx*(2sinx + 1) = 0 1) sinx = 0 x = πk, k∈z 2) 2sinx + 1 = 0 sinx = -1/2 x = -π/3 + 2πk, k∈z x = -2π/3 + 2πk, k∈z определим, при каких k корни уравнения принадлежат отрезку [5π/2; 4π] 5π/2 ≤ πk ≤ 4π 2.5 ≤ k ≤ 4, k∈z k = 3, 4 x1 = 3π; x2 = 4π 5π/2 ≤ -π/3 + 2πk ≤ 4π 17π/6 ≤ 2πk ≤ 13π/3 17/12 ≤ k ≤ 13/6, k∈z k = 2 x3 = -π/3 + 4π = 11π/3 5π/2 ≤ -2π/3 + 2πk ≤ 4π 19π/6 ≤ 2πk ≤ 14π/3 19/12 ≤ k ≤ 14/6, k∈z k = 2 x4 = -2π/3 + 4π = 10π/3 ответ: 10π/3; 11π/3; 3π; 4π
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