Решение
1.
1/cos²x = 2tg²x
1/cos²x = 2sin²x/cos²x
2sin²x = 1/2
sin²x = 1/4
1) sinx = - 1/2
x = (-1)^n arcsin(-1/2) + πk, k∈Z
x = (-1)^(n+1) arcsin(1/2) + πk, k∈Z
x = (-1)^(n+1) (π/6) + πk, k∈Z
2) sinx = 1/2
x = (-1)^n arcsin(1/2) + πk, k∈Z
x = (-1)^n (π/6) + πk, k∈Z
Ответ: x = (+ -)π/6 + πk, k∈Z
3.
2cos³x/5 + sin²x/5 = 1
2cos³x/5 + sin²x/5 = sin²x/5 + cos²x/5
2cos³x/5 - cos²x/5 = 0
cos²(x/5) *(2cosx/5 - 1) = 0
1) cos²(x/5) = 0
x/5 = π/2 + πn, n∈z
x = 5π/2 + 5πn, n∈Z
x = 5π/2(2n + 1)
2) 2cosx/5 - 1 = 0
cosx/5 = 1/2
x/5 = (+ -)arccos(1/2) + 2πk, k∈Z
x/5 = (+ -)(π/3) + 2πk, k∈Z
x = (+ -)(5π/3) + 10πk, k∈Z
x = [6k (+ -) 1]5π/3
Ответ: x = 5π/2 + 5πn, n∈Z ; x = (+ -)(5π/3) + 10πk, k∈Z