Помогите решить 2cosx/3+1=0
и еще
У=21/25-х(в квадрате)
Ответы на вопрос
Ответил nKrynka
0
2cos(x/3) + 1 = 0
cos(x/3) = - 1/2
x/3 = (+ -)arccos(-1/2) + 2πn, n∈Z
x/3 = (+ -)(π - arccos(1/2)) + 2πn, n∈Z
x/3 = (+ -)(π - π/3) + 2πn, n∈Z
x/3 = (+ -)(2π/3) + 2πn, n∈Z
x = (+ -)(2π) + 6πn, n∈Z
cos(x/3) = - 1/2
x/3 = (+ -)arccos(-1/2) + 2πn, n∈Z
x/3 = (+ -)(π - arccos(1/2)) + 2πn, n∈Z
x/3 = (+ -)(π - π/3) + 2πn, n∈Z
x/3 = (+ -)(2π/3) + 2πn, n∈Z
x = (+ -)(2π) + 6πn, n∈Z
Новые вопросы