Bonjour,
1) (2sin(x)+√3)(cos(x)+1)=0
Un produit de facteur est nul ssi l'un des facteurs est nul:
2sin(x)+√3=0
sin(x)=-√3/2
x=5π/6+2kπ
x=7π/6+2kπ
ou
cos(x)-1=0
cos(x)=-1
x=π+2kπ
Donc S={5π/6+2kπ, π+2kπ, 7π/6+2kπ}
2) 2cos²x-1=0
cos²x=1/2
cos(x)=-√2/2 ou cos(x)=√2/2
x=3π/4+2kπ
x=5π/4+2kπ
x=π/4+2kπ
x=7π/4+2kπ
d'où S={π/4+2kπ, 3π/4+2kπ, 5π/4+2kπ, x=7π/4+2kπ}
3) 2sin (x+π/4)=1
sin (x+π/4)=1/2
sin(x+π/4)=sin (π/6)
x+π/4=π/6+2kπ
x=π/6+2kπ-π/4
x=-π/12+2kπ
ou
x+π/4=π-π/4+2kπ
x=π-π/4+2kπ-π/4
x=π/2+2kπ
d'où S={-π/12+2kπ, π/2+2kπ}
4) sin(x)=cos(π/6)
sin(x)=sin(π/2-π/6)
sin(x)=sin(π/3)
x=π/6+2π
x=11π/6+2π
d'où S={π/6+2π, 11π/6+2π}
5)2sin²(x)+sin(x)-1=0
On pose t=sin(x) donc on a:
2t²+t-1=0
Δ=b²-4ac=1²+4(2)(-1)=9
t(1)=(-1+3)/4=1/2
t(2)=(-1-3)/4=-1
Comme t=sin(x) donc on a:
sin(x)=1/2
sin(x)=-1
x=5π/6
x=π/6
x=3π/2
D'où: S={π/6, 5π/6; 3π/2}
