Question

2sin^2 x+sin x-1=0


3cos^2 x-sin x+1=0

Answer 1

1
2sin²x+sinx-1=0
sinx=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒sinx=-1⇒x=-π/2+2πn,n∈z
a2=(-1+3)/4=1/2⇒sinx=1/2⇒x=(-1)^k*π/6+πk,k∈z
2
3cos²x-sinx+1=0
3-3sin²x-sinx+1=0
sinx=a
3a²+a-4=0
D=1+48=49
a1=(-1+7)/6=1⇒sinx=1⇒x=π/2+2πn,n∈z
a2=(-1-7)/6=-4/3⇒sinx=-4/3<-1 нет решения