Question

3 si 4
Dau coroana si 28 de puncte.
Materie clasa a 10-a.
Multumesc a3nticipat.​

Answer 1

Răspuns:

3.

3sinx+√3cosx=0║:√3

√3sinx+cosx=0

√3=tgπ/3 inlocuiesti in ecuatie

tgπ/3*sinx+cosx=0

(sinπ/3/cosπ/3)*sinx+cosx=0

sinπ/3*sinx+cosx*cosπ/3=0<=>

cosx*cosπ/3+sinπ/3*sinx=0

se aplica   formula cosαcosβ+sin αsinβ=cos(α-β)

cos(x-π/3)=0=>

x-π/3=π/2

x=π/2+π/3

x=5π/6

x=5π/6+2kπ

Explicație pas cu pas: