Вопрос школьника по предмету Алгебра
решите уравнение: 4cosx*cos2x*cos4x= sin8x
Ответ учителя по предмету Алгебра
Все решение сводится к использованию формулы
sin (2α) = 2 sin α · cos α
=============================
4cos(x) · cos(2x) · cos(4x) = sin(8x)
4cos(x) · cos(2x) · cos(4x) — 2sin(4x) · cos(4x) = 0
2cos(4x) · (2cos(x) · cos(2x) — sin(4x)) = 0
2cos(4x) · (2cos(x) · cos(2x) — 2sin(2x) · cos(2x)) = 0
4cos(4x) · cos(2x) · (cos(x) — sin(2x)) = 0
4cos(4x) · cos(2x) · (cos(x) — 2sin(x) · cos(x)) = 0
4cos(4x) · cos(2x) · cos(x) · (1 — 2sin(x)) = 0
Произведение равно нулю, когда один из множителей равен нулю.
1) cos (4x) = 0
4x = π/2 + πn
x₁ = π/8 + πn/4, n∈Z
2) cos(2x) = 0
2x = π/2 + πk
x₂ = π/4 + πk/2, k∈Z
3) cos(x) = 0
x₃ = π/2 + πm, m∈Z
4) 1 — 2sin(x) = 0
sin(x) = 1/2
x₄ = π/6 + 2πp, p∈Z
x₅ = 5π/6 + 2πq, q∈Z
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