3б)
Делим на cos²x≠0
3tg²x+2√3tgx+1=0
D=(2√3)²-4·3=12-12=0
tgx=-2√3/6;
tgx=-√3/3
x=(-π/6)+πk, k∈Z
О т в е т. (-π/6)+πk, k∈Z
5a)
делим на (1/√2)
(1/√2)sinx+(1/√2)cosx=(1/√2)
1/√2=sinπ/4=cosπ/4
(cos(π/4))sinx+(sin(π/4))cosx=1/√2
sin(x+π/4)=1/√2
x+(π/4)=π/4+2πk, k∈Z или х+(π/4)=3π/4+2πn, n∈Z
x=2πk, k∈Z или х=(π/2)+2πn, n∈Z
О т в е т.2πk; (π/2)+2πn; k, n∈Z
5б)
sin4x=2sin2x·cos2x
2cos²x=1+cos2x
1+cos2x+2sin2xcos2x=1
cos2x(1+2sin2x)=0
cos2x=0 или sin2x=-1/2
2x=(π/2)+πk, k∈Z; 2x= (-π/6)+2πn, n∈Z или 2х=(-5π/6)+2πm, m∈Z
x=(π/4)+(π/2)·k, k∈Z; x= (-π/12)+πn, n∈Z или х=(-5π/12)+πm, m∈Z
О т в е т.(π/4)+(π/2)·k; (-π/12)+πn; (-5π/12)+πm; k, n, m∈Z