1)1/(1-сos²x)=(cosx+3sinx)/sinx
1/sin²x=sinx(cosx+3sinx)/sin²x
1=sinx*cosx+3sin²x .sinx≠0
sin²x+cos²x-sinxcosx-3sin²x=0
-2sin²x-sinxcosx+cos²x=0 /-cos²x≠0
2tg²x+tgx-1=0
tgx=t. 2t²+t-1=0 D=1+4*2=9 t1=-1+3)/4=1/2 t2=(-1-3)4=-1
tgx=1/2 x=arctg1/2+πn. n∈Z
tgx=-1 x=-π/4+πk k∈Z