Решите уравнение 1/cos^2*x+3tgx-5=0
1/cos²x+3sinx/cosx-5=0 cosx≠0⇒x≠π/2+πk,k∈z 1+3sinxcosx-5cos²x=0 sin²x+cos²x+3sinxcosx-5cos²x=0/cos²x tg²x+3tgx-4=0 tgx=a a²+3a-4=0 a1+a2=-3 U a1*a2=-4 a1=-4⇒tgx=-4⇒x=-arctg4+πk,k∈z a2=1⇒tgx=1⇒x=π/4+πk,k∈z