Question

sa se rezolve in [0;2pi) ecuatia sinx+cosx=0

Answer 1

sinx=-cosx
Impartim ecuatia cu cosx si obtinem tgx=-1
x€{arctg(-1)+k*pi}
x€{-pi/4+k*pi}, k€Z
x=pi-(pi/4)=3pi/4 sau x=2pi-(pi/4)=7pi/4

Answer 2

de obicei ecuatiile cu sinx si cos x se rezolva imparind la cos x pt a ajunge la tg x

impartind cu cosx≠0, x≠(2k+1)π/2

avem

tgx+1=0

tgx=-1

x= kπ+arctg(-1), k∈Z

x= kπ+(-π/4)

in [0,2π) se gasesc x=3π/4 (pt k=1) si x=7π/4, pt k=2

altfel mai simplu

stim de la cl a 8-a ca tg45°=tg(π/4)=1

tg x este negativa in cadranele 2 si 4, unde obtinem x1=π/2+π/4=3π/4 si, respectiv, x2=2π-π/4=7π/4