Question

Решите неравенство: sinx + cos2x > 1.

Answer 1

Решение
sinx + cos2x > 0 
sinx + 1 - 2sin^2(x) > 0
2sin^2(x) - sinx  - 1 <  0
sinx = z
2*z^2 - z - 1 = 0
D = 1 + 4*1*1 = 9
z1 = (1 - 3)/4
 z1 = -1/2
z2 = (1 + 3)/4
z2 = 1
1)  sinx = -1/2
x = (-1)^(n)arcsin(-1/2) + πn, n∈Z
x = (-1)^(n+1)arcsin(1/2) + πn, n∈Z
x1 = (-1)^(n+1)(π/6) + πn, n∈Z
2)  sinx = 1
x2 = π/2 = 2πk, k∈Z
Ответ: x1 = (-1)^(n+1)(π/6) + πn, n∈Z;  x2 = π/2 = 2πk, k∈Z.