Не могу решить, , cos2x/3+5sinx/3+2=0

Cos(2x/3) +5sin(x/3)+2=0
1-2sin²(x/3)+5sin(x/3)+2=0
2sin²(x/3)-5sin(x/3)-3=0
sin(x/3)=a
2a²-5a-3=0
D=25+24=49
a1=(5-7)/4=-1/2⇒sin(x/3)=-1/2⇒(x/3)=(-1)^n+1 * π/6+πn⇒x=(-1)^n+1 * π/2+3πn
a2=(5+7)/4=3⇒sin(x/3)=3∈[-1;1]