4(1 - Cos²x) - Cosx -1 = 0
4 - 4Cos²x - Cosx -1 = 0
4Cos²x + Cosx -3 = 0
Cosx = t
4t² + t - 3 = 0
D = 49
a) t₁ = (-1+7)/8 = 6/8 = 3/4 б) t₂ = (-1 - 7)/8 = -1
Cosx = 3/4 Сos x = -1
x = +-arcCos(3/4) + 2πk , k ∈Z x = π + 2πn , n ∈ Z
F(2x)=1+8x^3
f(x/2)=1+x^3/8
(f(2x)-1)(f(x/2)-1)+=(1+8x^3-1)(1+x^3/8-1)+1=(8x^3/8)*x^3=x^6+1
(f(x)-1)^2+1=(1+x^3-1)^2+1=x^6+1
161+1+1+1+1 во так наверное
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