Question

решите неравенство (2/tgx+1) < 2-tgx

Answer 1

2/(tgx+1)<2-tgx
tgx=a
2/(a+1)-2-a<0
(2-2a+2+a²+a)/(a+1)<0
(a²-a+4)/(a+1)<0
a²-a+4>0 при любом а,т.к. D=-15<0⇒a+1<0⇒a<-1
tgx<-1
x∈(-π/2+πk;-π/4+πk,k∈z)