Răspuns :
Domeniul de existenta al ecuatiei
x≠(2k+1)π/2 -in aceste puncte tangenta nu este definita
cosx*(sinx/cosx*tgx-1)=0
cosx*(tg²x-1)=0=>
cosx=0 si (tg²x-1)=0
c0sx=0 => x=(2k+1)π/2∉domeniului de existenta stabilit anterior
tg²x-1=0=>
tg²x=1
tgx=+/1
tgx=1=>x=π/4
solutia generala
x1=π/4+/-kπ k∈Z
tgx=-1
x=5π/4
x2=5π/4+/-kπ
x={π/4+/-kπ}U{5π/4+/-kπ}={{π/4+/-kπ}
x≠(2k+1)π/2 -in aceste puncte tangenta nu este definita
cosx*(sinx/cosx*tgx-1)=0
cosx*(tg²x-1)=0=>
cosx=0 si (tg²x-1)=0
c0sx=0 => x=(2k+1)π/2∉domeniului de existenta stabilit anterior
tg²x-1=0=>
tg²x=1
tgx=+/1
tgx=1=>x=π/4
solutia generala
x1=π/4+/-kπ k∈Z
tgx=-1
x=5π/4
x2=5π/4+/-kπ
x={π/4+/-kπ}U{5π/4+/-kπ}={{π/4+/-kπ}