Предмет: Алгебра,
автор: Qwer139
Решите уравнение: cos^2 x(x+pi/3)=1
Ответы
Автор ответа:
0
sin²(x + π/3) = 1
1) sin(x + π/3) = - 1
x + π/3 = - π/2 + 2πk, k ∈z
x = - π/3 - π/2 + 2πk, k ∈z
x = - 5π/6 + 2πk, k ∈z
2) sin(x + π/3) = 1
x + π/3 = π/2 + 2πn, n∈Z
x= - π/3 + π/2 + 2πn, n∈Z
x= π/6 + 2πn, n∈Z
1) sin(x + π/3) = - 1
x + π/3 = - π/2 + 2πk, k ∈z
x = - π/3 - π/2 + 2πk, k ∈z
x = - 5π/6 + 2πk, k ∈z
2) sin(x + π/3) = 1
x + π/3 = π/2 + 2πn, n∈Z
x= - π/3 + π/2 + 2πn, n∈Z
x= π/6 + 2πn, n∈Z
Автор ответа:
0
cos²(x + π/3) = 1
1) cos(x + π/3) = - 1
x + π/3 = π + 2πk, k ∈z
x = - π/3 + π + 2πk, k ∈z
x = 2π/3 + 2πk, k ∈z
2) cos(x + π/3) = 1
x + π/3 = 2πn, n∈Z
x= - π/3 + 2πn, n∈Z
1) cos(x + π/3) = - 1
x + π/3 = π + 2πk, k ∈z
x = - π/3 + π + 2πk, k ∈z
x = 2π/3 + 2πk, k ∈z
2) cos(x + π/3) = 1
x + π/3 = 2πn, n∈Z
x= - π/3 + 2πn, n∈Z
Автор ответа:
0
cos²(x+π/3)=1
[1+cos(2x+2π/3)]/2=1
1+cos(2x+2π/3)=2
cos(2x+2π/3)=1
2x+2π/3=2πn
2x=-2π/3+2πn
x=-π/3+πn
[1+cos(2x+2π/3)]/2=1
1+cos(2x+2π/3)=2
cos(2x+2π/3)=1
2x+2π/3=2πn
2x=-2π/3+2πn
x=-π/3+πn
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