Question

Cos^2x=sin^2(x/2). 13 егэ профиль

Answer 1

Cos²x=Sin²(x/2)
(1 - 2Sin²x/2)² -Sin²x/2 = 0
1 - 4Sin²x/2 + 4Sin⁴x/2 - Sin²x/2 = 0
4Sin⁴x/2 -5Sin²x/2 +1 = 0
Sin²x/2 = t
4t² -5t +1 = 0
D = 25 -16 = 9
t₁ = 1                                   t₂=1/4
Sin²x/2 = 1                        Sin²x/2 = 1/4
Sinx/2 = +-1                      Sinx/2 = +-1/2
а)Sinx/2 = 1
    x/2 = π/2 +2πk , k ∈Z
    x = π +4πk , k ∈Z 
б) Sin x/2 = -1
    x/2 = -π/2 +2πn , n∈Z
    x = -π +4πn , n∈Z 
в) Sinx/2 = 1/2
x/2 = (-1)^m *π/6 + mπ, m ∈Z
x = (-1)^m*π/3 +2πm , m∈Z
г) Sinx/2 = -1/2
x/2 = (-1)^(c+1)*π/6 + cπ, c ∈ Z
x = (-1)^(c+1)*π/3 + 2cπ, c∈Z