Question

(sinx+cosx)/(cos2x)>0

Answer 1

(sinx+sin(π/2+x)/cos2x>0
2sin((x+π/2+x)/2)cos((x-π/2-x)/2)/cos2x>0
2sin(x+π/4)cos(-π/4)/cos2x>0 /:2
sin(x+π/4)cosπ/4/cos2x>0
sin(x+π/4)>0 i cos2x>0)     lub ( sin(x+π/4)<0 i cos2x<0)
(x+π/4)∈(0,π) i    2x∈(-π/2,π/2)   lub      (x+π/4∈(-π,0) i 2x∈(-π,-π/2)U(π/2,π)
x∈(-π/4,3/4π) i x∈(-π/4,π/4)     lub    x∈(-5/4π,-π/4)   i  x∈(-π/2,-π/4)U(π/4,π/2)
x∈(-π/4,π/4)                  lub  x∈(-π/2,-π/4)
x∈(-π/2,-π/4)U(-π/4,π/4)