Answer:
Explanation:
\[2~ sin^2 x+3sin x+1=0\]
\[2sin^2x+2sin x+sin x+1=0\]
2sinx(sin x+1)+1(sin x+1)=0
(sin x+1)(2 sin x+1)=0
either sin x+1=0
sin x=-1=sin 3π/2=sin (2nπ+3π/2)
x=2nπ+3π/2,where n is an integer.
or 2sin x+1=0
sin x=-1/2=-sin π/6=sin (π+π/6),sin (2π-π/6)=sin (2nπ+7π/6),sin (2nπ+11π/6)
x=2nπ+7π/6,2nπ+11π/6,
where n is an integer.