Решение
(2,5sinx*cosx + 1) / cosx = 0
(5/4 sin2x + 1)(cosx) = 0
1) 5/4 sin2x + 1 = 0
5/4 sin2x = - 1
sin2x = - 4/5
2x = (-1)^narcsin(-4/5) + πn, n∈Z
2x = (-1)^(n+1)arcsin(4/5) + πn, n∈Z
x =(1/2)* (-1)^(n+1)arcsin(4/5) + (1/2)*πn, n∈Z
2) cosx = 0
x = π/2 + πk, k∈Z