Question

Cos4x+2cos^2x=1хеееелппжжжж​

Answer 1

Cos4x + 2Cos²x = 1

Cos4x = 1 - 2Cos²x

2Cos²2x -1 = - Cos2x

Cos2x = t

2t² +t -1 = 0

D = 1 -4*2*(-1)=9

t₁ = 1/2                                          t₂=-1

Cos2x = 1/2                             Cos2x = -1

2x = +-π/3 + 2πk, k ∈Z              2x = π + 2πn , n ∈ Z

x = +- π/6 +πk , k ∈Z                   x = π/2 + πn , n ∈Z

(использовалась формула:  Сos2α= 1 - 2Cos²α  )

Answer 2

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