Question

Решите уравнение 1/cos^2*x+3tgx-5=0

Answer 1

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