exo 34 comment on resoud

1) Sin(-x+3π/2)-sin(2x+π/3)=0
⇔sin(-x+3π/2)=sin(2x+π/3)
2 angles α et β ont même sinus si et seulement si α=β ou α=π-β
Donc soit -x+3π/2=2x+π/3
3x=3π/2-π/3=9π/6-2π/6=7π/6
x=7π/18

Soit -x+3π/2=π-2x-π/3
x=π-π/3-3π/2=6π/6-2π/6-9π/6=-5π/6

Donc S={7π/18+2kπ;-5π/6+2kπ}

2) sin(5x-5π/6)+sin(2x+π/4)=0
⇔sin(5x-5π/6)=-sin(2x+π/4)
Or -sinα=sin(-α) donc
sin(5x-5π/6)=sin(-2x-π/4)
Soit 5x-5π/6=-2x-π/4
7x=5π/6-π/4=10π/12-3π/12=7π/12
x=π/12

Soit 5x-5π/6=π+2x+π/4
3x=5π/6+π+π/4=10π/12+12π/12+3π/12=25π/12
x=25π/36

S={π/12+2kπ;25π/36+2kπ}