Помогите пожалуйста, завтра сдавать))). Created by marenkovartyom. algebra-ru

Помогите пожалуйста, завтра сдавать)))...

October 2021 1 11 Report

Помогите пожалуйста, завтра сдавать)))

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nKrynka Решение
√2cos²x = sin(π/2 + x)
√2cos²x - cosx = 0
cosx*(√2cosx - 1) = 0
1) cosx = 0
x = π/2 + πn, n∈Z
2) √2cosx - 1 = 0
cosx = 1/√2
x = (+ -)arccos(1/√2) + 2πk, k∈Z
x = (+ -)(π/4) + 2πk, k∈Z
- 7π/2 ≤ π/4 + 2πk ≤ - 2π
- 7π/2 - π/4 ≤ 2πk ≤ - 2π - π/4
-15π/4 ≤ 2πk ≤ - 9π/4
15/8 ≤ k ≤ -9π/8
- 7π/2 ≤ - π/4 + 2πk ≤ - 2π
- 7π/2 + π/4 ≤ 2πk ≤ - 2π + π/4
- 13π/4 ≤ 2πk ≤ - 7π/4
- 13/8 ≤ k ≤ - 7/8
- 1 (5/8) ≤ k ≤ - 7/8
отсюда k = - 1
Значит, x = (+ -)(π/4) - 2π
x₁ = - π/4 - 2π = - 9π/4
x₂ = π/4 - 2π = - 7π/4

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