cos^2x-sin^2x-sinx=0...

mrrodich @mrrodich

August 2022 1 6 Report

cos^2x-sin^2x-sinx=0

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nKrynka Решение
Cos²x - sin²x - sinx = 0
1 - sin²x - sin²x - sinx = 0
2sin²x + sinx - 1 = 0
sinx = t
2t² + t - 1 = 0
D = 1 + 4*2*1 = 9
t₁ = (- 1 - 3)/4
t₁ = - 1
t₂ = (- 1 + 3)/4
t₂ = 1/2
1) sinx = - 1
x₁ = - π/2 + 2πk, k∈Z
2) sinx = 1/2
x = (-1)^n *arcsin(1/2) + πn, n∈ Z
x₂ = (-1)^n *(π/6) + πn, n∈ Z

3 votes Thanks 6

LordByron Чёрт,ты прав

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